Module 6: Electromagnetism | Year 12 Physics HSC | StudyCave

Pillar 1 of 3

Charged Particles, Conductors & Fields

Explore how charged particles and current-carrying conductors interact with electric and magnetic fields through fundamental force equations.

Electric Fields

An electric field is a region of space where a charged particle experiences an electric force. The field strength (intensity) describes the force per unit charge at any point in the field.

Key Equations

Electric Force on a Charge:

F = qE

Where:

$F$ = electric force (N)
$q$ = charge (C)
$E$ = electric field strength (N/C or V/m)

Field Strength Between Parallel Plates:

E = \frac{V}{d}

Where:

$E$ = electric field strength (V/m)
$V$ = potential difference between plates (V)
$d$ = separation distance between plates (m)

Key Insight: Between parallel plates, the electric field is uniform (constant magnitude and direction). Field lines point from positive to negative plate.

Important Properties:

Direction of Force

Positive charges experience force in the direction of the field. Negative charges experience force opposite to the field direction.

Uniform Fields

Parallel plates create a uniform electric field in the region between them. Outside this region, the field becomes non-uniform.

Field Strength Units

Electric field strength can be expressed as N/C (force per charge) or V/m (voltage gradient). These are equivalent: $1 \text{ N/C} = 1 \text{ V/m}$ .

Independence of Charge

The field strength $E = V/d$ depends only on the voltage and plate separation, not on the test charge placed in the field.

Worked Example

Two parallel plates are separated by 5.0 cm and have a potential difference of 2000 V across them. An electron ( $q = -1.6 \times 10^{-19} \text{ C}$ ) is placed between the plates. Calculate: (a) the electric field strength, (b) the force on the electron, (c) the direction of the force.

Given:

$d = 5.0 \text{ cm} = 0.050 \text{ m}, \quad V = 2000 \text{ V}, \quad q = -1.6 \times 10^{-19} \text{ C}$

(a) Electric field strength:

$E = \frac{V}{d}$

$E = \frac{2000}{0.050}$

$E = \mathbf{40,000 \text{ V/m}}$

(b) Force on the electron:

$F = qE$

$F = (1.6 \times 10^{-19})(40,000)$

$F = \mathbf{6.4 \times 10^{-15} \text{ N}}$

The electron has negative charge, so it experiences force opposite to the electric field direction.

If the field points from positive to negative plate (downward), the electron is pushed upward toward the positive plate.

Practice Question

A proton ( $q = +1.6 \times 10^{-19} \text{ C}$ ) experiences a force of $3.2 \times 10^{-14} \text{ N}$ when placed in a uniform electric field. Calculate the electric field strength. If this field exists between two plates 8.0 cm apart, what is the potential difference across the plates?

Show Answer

Step 1: Calculate field strength

$F = qE \implies E = \frac{F}{q}$

$E = \frac{3.2 \times 10^{-14}}{1.6 \times 10^{-19}}$

$E = \mathbf{2.0 \times 10^5 \text{ V/m}}$

Step 2: Calculate potential difference

$E = \frac{V}{d} \implies V = Ed$

$V = (2.0 \times 10^5)(0.080)$

$V = \mathbf{16,000 \text{ V}}$

Work Done in Electric Fields

When a charged particle moves through an electric field, work is done either by the field (if moving with the force) or against the field (if moving opposite to the force). This work transfers energy to or from the particle.

Work Done Equations

Work Done Using Voltage:

W = qV

Where:

$W$ = work done (J)
$q$ = charge (C)
$V$ = potential difference (V)

Work Done Using Force:

W = Fd

Where:

$W$ = work done (J)
$F$ = electric force (N)
$d$ = displacement in direction of force (m)

Equivalence: Since $F = qE$ and $E = V/d$ , we can show: $W = Fd = qEd = q(V/d) \times d = qV$

Energy Transformations:

Accelerating Charge

When a charge moves in the direction of the electric force, work is done by the field. Electrical potential energy decreases, kinetic energy increases: $qV = \frac{1}{2}mv^2$

Decelerating Charge

When a charge moves opposite to the electric force, work is done against the field. Kinetic energy decreases, electrical potential energy increases.

Sign Convention

Work done by the field is positive when the force and displacement are in the same direction. Work done against the field is negative (or requires external work input).

The Electron Volt (eV)

One electron volt is the energy gained by an electron accelerating through 1 V: $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$

Worked Example

An electron (mass $m = 9.1 \times 10^{-31} \text{ kg}$ , charge $q = -1.6 \times 10^{-19} \text{ C}$ ) is accelerated from rest through a potential difference of 5000 V. Calculate: (a) the work done on the electron, (b) the final speed of the electron.

Given:

$m = 9.1 \times 10^{-31} \text{ kg}, \quad q = 1.6 \times 10^{-19} \text{ C}, \quad V = 5000 \text{ V}, \quad u = 0$

(a) Work done on the electron:

$W = qV$

$W = (1.6 \times 10^{-19})(5000)$

$W = \mathbf{8.0 \times 10^{-16} \text{ J}}$

(Using magnitude of charge)

(b) Final speed:

Work done equals kinetic energy gained:

$W = \frac{1}{2}mv^2$

$8.0 \times 10^{-16} = \frac{1}{2}(9.1 \times 10^{-31})v^2$

$v^2 = \frac{2 \times 8.0 \times 10^{-16}}{9.1 \times 10^{-31}}$

$v^2 = 1.76 \times 10^{15}$

$v = \mathbf{4.2 \times 10^7 \text{ m/s}}$

(About 14% the speed of light!)

Practice Question

A proton enters a uniform electric field with a speed of $2.0 \times 10^6 \text{ m/s}$ and comes to rest after traveling 15 cm. The mass of a proton is $1.67 \times 10^{-27} \text{ kg}$ . Calculate: (a) the work done on the proton, (b) the potential difference through which it traveled.

Show Answer

(a) Work done on proton:

The proton loses all its kinetic energy, so work done equals initial KE:

$W = \frac{1}{2}mu^2 - \frac{1}{2}mv^2$

$W = \frac{1}{2}(1.67 \times 10^{-27})(2.0 \times 10^6)^2 - 0$

$W = \frac{1}{2}(1.67 \times 10^{-27})(4.0 \times 10^{12})$

$W = \mathbf{-3.34 \times 10^{-15} \text{ J}}$

(Negative because field does negative work on the proton)

(b) Potential difference:

$W = qV \implies V = \frac{W}{q}$

$V = \frac{3.34 \times 10^{-15}}{1.6 \times 10^{-19}}$

$V = \mathbf{20,875 \text{ V}}$

Trajectory Analysis in Electric Fields

When a charged particle enters a uniform electric field perpendicular to the field lines, it follows a parabolic trajectory similar to projectile motion. The motion can be analyzed by separating it into horizontal and vertical components.

Motion Analysis

Setup: Consider a charged particle entering between parallel plates with initial horizontal velocity $v_0$ . The electric field is vertical (perpendicular to initial velocity).

Horizontal Motion

• No horizontal force
• $a_x = 0$
• $v_x = v_0$ (constant)
• $x = v_0 t$

Vertical Motion

• Electric force: $F = qE$
• $a_y = \frac{qE}{m}$ (constant)
• $v_y = a_y t$
• $y = \frac{1}{2}a_y t^2$

Trajectory Equation:

Eliminate time by using $t = x/v_0$ :

y = \frac{1}{2}a_y \left(\frac{x}{v_0}\right)^2 = \frac{qE}{2mv_0^2}x^2

This is a parabola in the form $y = kx^2$ , identical in form to projectile motion under gravity.

Key Insights:

Analogy with Gravity

The electric field acts like "gravity" in the vertical direction. Just as projectiles follow parabolic paths under gravity, charged particles follow parabolic paths in uniform electric fields.

Deflection Direction

Positive charges deflect toward the negative plate (in the field direction). Negative charges deflect toward the positive plate (opposite the field direction).

Mass Dependence

Lighter particles (like electrons) experience greater deflection than heavier particles (like protons) with the same charge and initial velocity: $a \propto 1/m$

CRO Application

Cathode Ray Oscilloscopes use this principle. Electrons are deflected vertically by electric fields between parallel plates, creating the display pattern.

Worked Example

An electron enters a uniform electric field between two parallel plates. The plates are 10 cm long and separated by 2.0 cm. The potential difference across the plates is 400 V. The electron enters horizontally with a speed of $5.0 \times 10^6 \text{ m/s}$ . Calculate: (a) the electric field strength, (b) the vertical acceleration, (c) the vertical deflection as it exits the plates.

Given: $m_e = 9.1 \times 10^{-31} \text{ kg}, \quad e = 1.6 \times 10^{-19} \text{ C}$

Given:

$L = 0.10 \text{ m}, \quad d = 0.020 \text{ m}, \quad V = 400 \text{ V}, \quad v_0 = 5.0 \times 10^6 \text{ m/s}$

(a) Electric field strength:

$E = \frac{V}{d} = \frac{400}{0.020} = \mathbf{20,000 \text{ V/m}}$

(b) Vertical acceleration:

$F = eE = ma$

$a = \frac{eE}{m} = \frac{(1.6 \times 10^{-19})(20,000)}{9.1 \times 10^{-31}}$

$a = \frac{3.2 \times 10^{-15}}{9.1 \times 10^{-31}} = \mathbf{3.52 \times 10^{15} \text{ m/s}^2}$

Time in field: $t = \frac{L}{v_0} = \frac{0.10}{5.0 \times 10^6} = 2.0 \times 10^{-8} \text{ s}$

Vertical displacement:

$y = \frac{1}{2}at^2 = \frac{1}{2}(3.52 \times 10^{15})(2.0 \times 10^{-8})^2$

$y = \frac{1}{2}(3.52 \times 10^{15})(4.0 \times 10^{-16})$

$y = \mathbf{7.04 \times 10^{-4} \text{ m} = 0.70 \text{ mm}}$

The electron deflects by 0.70 mm toward the positive plate as it travels through the field.

Practice Question

A proton enters a uniform electric field of strength $1.0 \times 10^4 \text{ V/m}$ with an initial horizontal velocity of $2.0 \times 10^5 \text{ m/s}$ . How long does it take for the proton's vertical velocity component to equal its horizontal velocity component? (Mass of proton: $1.67 \times 10^{-27} \text{ kg}$ )

Show Answer

Solution:

Vertical acceleration: $a = \frac{eE}{m} = \frac{(1.6 \times 10^{-19})(1.0 \times 10^4)}{1.67 \times 10^{-27}}$

$a = 9.58 \times 10^{11} \text{ m/s}^2$

We want $v_y = v_x$ :

$v_y = at = v_0$

$t = \frac{v_0}{a} = \frac{2.0 \times 10^5}{9.58 \times 10^{11}}$

$t = \mathbf{2.09 \times 10^{-7} \text{ s} = 209 \text{ ns}}$

Magnetic Forces on Particles

When a charged particle moves through a magnetic field, it experiences a force perpendicular to both its velocity and the magnetic field. This force causes the particle to follow a circular or helical path.

Magnetic Force on Moving Charges

F = qvB\sin\theta

Where:

$F$ = magnetic force (N)
$q$ = charge (C)
$v$ = velocity of the particle (m/s)
$B$ = magnetic field strength (T)
$\theta$ = angle between velocity vector and magnetic field

Key Insight: The force is always perpendicular to both $\vec{v}$ and $\vec{B}$ . Maximum force occurs when $\theta = 90°$ (perpendicular), and zero force when $\theta = 0°$ or $180°$ (parallel).

Critical Properties:

Force is Perpendicular

The magnetic force is always perpendicular to the particle's velocity. This means it does NO work on the particle (since $W = Fd\cos\theta$ and $\theta = 90°$ ). The force only changes direction, not speed.

Circular Motion

When a charged particle enters a uniform magnetic field perpendicularly, the perpendicular force provides centripetal acceleration, causing the particle to move in a circular path.

Speed Remains Constant

Since magnetic force does no work, kinetic energy (and therefore speed) remains constant. Only the direction of velocity changes, maintaining circular motion at constant speed.

Charge Sign Matters

Positive and negative charges curve in opposite directions in the same magnetic field. This principle is used in mass spectrometers to separate particles.

Radius of Circular Path

Derivation:

For circular motion, magnetic force provides centripetal force:

F_{\text{magnetic}} = F_{\text{centripetal}}

qvB = \frac{mv^2}{r}

Solving for radius:

r = \frac{mv}{qB}

Period of circular motion:

T = \frac{2\pi r}{v} = \frac{2\pi m}{qB}

Note: Period is independent of velocity!

Worked Example

An electron ( $m = 9.1 \times 10^{-31} \text{ kg}, q = 1.6 \times 10^{-19} \text{ C}$ ) enters a uniform magnetic field of 0.050 T at a speed of $2.0 \times 10^6 \text{ m/s}$ perpendicular to the field. Calculate: (a) the magnetic force on the electron, (b) the radius of its circular path, (c) the period of one complete circle.

Given:

$m = 9.1 \times 10^{-31} \text{ kg}, \quad q = 1.6 \times 10^{-19} \text{ C}$

$v = 2.0 \times 10^6 \text{ m/s}, \quad B = 0.050 \text{ T}, \quad \theta = 90°$

(a) Magnetic force:

$F = qvB\sin\theta = qvB\sin(90°)$

$F = (1.6 \times 10^{-19})(2.0 \times 10^6)(0.050)(1)$

$F = \mathbf{1.6 \times 10^{-14} \text{ N}}$

(b) Radius of circular path:

$r = \frac{mv}{qB}$

$r = \frac{(9.1 \times 10^{-31})(2.0 \times 10^6)}{(1.6 \times 10^{-19})(0.050)}$

$r = \frac{1.82 \times 10^{-24}}{8.0 \times 10^{-21}}$

$r = \mathbf{2.28 \times 10^{-4} \text{ m} = 0.228 \text{ mm}}$

$T = \frac{2\pi m}{qB}$

$T = \frac{2\pi (9.1 \times 10^{-31})}{(1.6 \times 10^{-19})(0.050)}$

$T = \frac{5.71 \times 10^{-30}}{8.0 \times 10^{-21}}$

$T = \mathbf{7.14 \times 10^{-10} \text{ s} = 0.714 \text{ ns}}$

Practice Question

A proton (mass $1.67 \times 10^{-27} \text{ kg}$ ) and an electron travel at the same speed through the same magnetic field perpendicular to their motion. Compare the radii of their circular paths. Which particle follows the larger circle?

Show Answer

Analysis:

The radius formula is: $r = \frac{mv}{qB}$

Both particles have the same charge magnitude ( $q = 1.6 \times 10^{-19} \text{ C}$ ), same speed $v$ , and same field $B$ .

Therefore: $r \propto m$

Mass ratio: $\frac{m_p}{m_e} = \frac{1.67 \times 10^{-27}}{9.1 \times 10^{-31}} \approx 1836$

$\frac{r_p}{r_e} = \frac{m_p}{m_e} \approx 1836$

Answer: The proton follows a circle approximately 1836 times larger in radius than the electron. Heavier particles require larger radii to maintain the same speed in circular motion.

The Motor Effect (Force on Wires)

When a current-carrying conductor is placed in a magnetic field, it experiences a force. This is called the motor effect and is the principle behind electric motors. The force arises because moving charges (current) in the wire experience magnetic forces.

Force on a Current-Carrying Conductor

F = BIL\sin\theta

Where:

$F$ = force on the conductor (N)
$B$ = magnetic field strength (T)
$I$ = current through the conductor (A)
$L$ = length of conductor in the field (m)
$\theta$ = angle between current direction and magnetic field

Maximum Force: When the current is perpendicular to the magnetic field ( $\theta = 90°$ ), the force is maximum: $F = BIL$

Zero Force: When the current is parallel to the magnetic field ( $\theta = 0°$ or $180°$ ), the force is zero.

Understanding the Motor Effect:

Origin of Force

Current is moving charges. Each charge experiences $F = qvB$ . With billions of charges moving, the total force on the wire is $F = BIL$ .

Direction of Force

The force is perpendicular to both the current direction and the magnetic field. Use the Right Hand Palm Rule to determine the direction.

Applications

Electric motors, loudspeakers, electromagnetic rail guns, and meters all use the motor effect to convert electrical energy to mechanical energy.

Reversing Force

Reversing either the current direction OR the magnetic field direction reverses the force. Reversing both keeps the force in the same direction.

Worked Example

A straight wire of length 25 cm carries a current of 8.0 A. It is placed perpendicular to a uniform magnetic field of strength 0.40 T. Calculate: (a) the force on the wire, (b) the force if the wire makes an angle of 30° with the field.

Given:

$L = 25 \text{ cm} = 0.25 \text{ m}, \quad I = 8.0 \text{ A}, \quad B = 0.40 \text{ T}$

(a) Force when perpendicular ( $\theta = 90°$ ):

$F = BIL\sin\theta = BIL\sin(90°)$

$F = (0.40)(8.0)(0.25)(1)$

$F = \mathbf{0.80 \text{ N}}$

(b) Force at 30° to field:

$F = BIL\sin\theta = BIL\sin(30°)$

$F = (0.40)(8.0)(0.25)(0.5)$

$F = \mathbf{0.40 \text{ N}}$

The force is exactly half the maximum value.

Practice Question

A horizontal wire of length 0.50 m carrying a current of 15 A is placed in a vertical magnetic field of 0.30 T. The wire experiences a horizontal force of 1.5 N. Calculate the angle between the wire and the magnetic field.

Show Answer

Solution:

Using $F = BIL\sin\theta$ :

$1.5 = (0.30)(15)(0.50)\sin\theta$

$1.5 = 2.25\sin\theta$

$\sin\theta = \frac{1.5}{2.25} = 0.667$

$\theta = \arcsin(0.667) = \mathbf{41.8°}$

Force Between Parallel Wires

Two parallel current-carrying wires exert magnetic forces on each other. Wires with currents in the same direction attract, while wires with currents in opposite directions repel.

Force Per Unit Length Between Parallel Wires

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r}

Where:

$F/L$ = force per unit length (N/m)
$\mu_0 = 4\pi \times 10^{-7} \text{ Tm/A}$ (permeability of free space)
$I_1, I_2$ = currents in the two wires (A)
$r$ = perpendicular distance between the wires (m)

Note: This formula gives the magnitude of force. The direction depends on current directions.

Why Parallel Wires Interact

Step 1: Wire 1 creates a magnetic field around itself (circular field lines using right-hand grip rule).

Step 2: Wire 2 sits in this magnetic field created by Wire 1.

Step 3: Wire 2 (carrying current $I_2$ ) experiences a force due to being in Wire 1's magnetic field (motor effect: $F = BIL$ ).

Step 4: By Newton's Third Law, Wire 1 experiences an equal and opposite force from Wire 2.

Same Direction → Attract

When currents flow in the same direction, the magnetic field between the wires is weakened. The stronger field on the outside pushes the wires together.

Opposite Direction → Repel

When currents flow in opposite directions, the magnetic field between the wires is strengthened. The wires are pushed apart.

Worked Example

Two long parallel wires are separated by a distance of 10 cm. Wire 1 carries a current of 20 A and Wire 2 carries a current of 15 A in the same direction. Calculate: (a) the force per unit length on each wire, (b) the total force on a 2.5 m length of wire, (c) whether the force is attractive or repulsive.

Given:

$I_1 = 20 \text{ A}, \quad I_2 = 15 \text{ A}, \quad r = 10 \text{ cm} = 0.10 \text{ m}$

$\mu_0 = 4\pi \times 10^{-7} \text{ Tm/A}$

(a) Force per unit length:

$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r}$

$\frac{F}{L} = \frac{4\pi \times 10^{-7} \times 20 \times 15}{2\pi \times 0.10}$

$\frac{F}{L} = \frac{4 \times 10^{-7} \times 300}{2 \times 0.10}$

$\frac{F}{L} = \frac{1.2 \times 10^{-4}}{0.2}$

$\frac{F}{L} = \mathbf{6.0 \times 10^{-4} \text{ N/m}}$

(b) Total force on 2.5 m length:

$F = \frac{F}{L} \times L = (6.0 \times 10^{-4})(2.5)$

$F = \mathbf{1.5 \times 10^{-3} \text{ N} = 1.5 \text{ mN}}$

The currents flow in the same direction, therefore the wires attract each other.

Practice Question

Two parallel wires 5.0 cm apart carry currents in opposite directions. If the force per unit length between them is $8.0 \times 10^{-4} \text{ N/m}$ , and one wire carries 25 A, what current flows in the other wire?

Show Answer

Solution:

Using $\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r}$ :

$8.0 \times 10^{-4} = \frac{4\pi \times 10^{-7} \times 25 \times I_2}{2\pi \times 0.05}$

$8.0 \times 10^{-4} = \frac{4 \times 10^{-7} \times 25 \times I_2}{2 \times 0.05}$

$8.0 \times 10^{-4} = \frac{10^{-5} I_2}{0.1}$

$8.0 \times 10^{-4} = 10^{-4} I_2$

$I_2 = \mathbf{8.0 \text{ A}}$

Since currents are in opposite directions, the force is repulsive.

Hand Rules for Electromagnetic Forces

Hand rules are essential tools for determining the direction of forces, fields, and currents in electromagnetic situations. The Right Hand Palm Rule (also called the Right Hand Slap Rule) is the most commonly used for finding force directions.

Right Hand Palm Rule (Motor Effect)

Used for: Finding the direction of force on a current-carrying wire or moving charge

How to use:

Fingers point in the direction of the magnetic field ( $\vec{B}$ )
Thumb points in the direction of the current or velocity ( $\vec{I}$ or $\vec{v}$ for positive charges)
Palm pushes in the direction of the force ( $\vec{F}$ )

Memory aid: "FBI" - Fingers = B-field, thuMb = I (current/motion), Palm = Force

For Negative Charges (electrons):

Use the same rule but point your thumb opposite to the direction of electron motion, OR use your left hand instead of right hand.

Fleming's Left Hand Rule (Alternative)

Used for: Same purpose as Right Hand Palm Rule

How to use (LEFT hand):

First finger (index) = Field direction ( $\vec{B}$ )
seCond finger (middle) = Current direction ( $\vec{I}$ )
thuMb = Motion/Force direction ( $\vec{F}$ )

Memory aid: "FBI" again - First finger = Field, seCond = Current, thuMb = Motion

Note: Both rules give the same result. Choose whichever you find easier to remember and use consistently.

Right Hand Grip Rule (Magnetic Fields)

Used for: Finding magnetic field direction around a current-carrying wire

For a straight wire:

Thumb points in the direction of conventional current
Fingers curl around the wire in the direction of the magnetic field lines

For a solenoid (coil):

Fingers curl in the direction of current flow around the coil
Thumb points toward the North pole of the electromagnet (direction of field inside)

Remember: This rule is for finding the field created by a current, not the force on the current.

Common Applications:

Motors

Use Right Hand Palm Rule to find the force direction on each side of the motor coil. This tells you which direction the motor will rotate.

Charged Particle Deflection

Use Right Hand Palm Rule with thumb pointing in velocity direction to find which way a positive charge will curve in a magnetic field.

Parallel Wire Interaction

Use Right Hand Grip to find the field from one wire, then Right Hand Palm to find the force on the second wire in that field.

Solenoids

Use Right Hand Grip for coils to identify which end is the North pole of an electromagnet based on current direction.

Practice Question

A horizontal wire carries current from west to east through a vertical magnetic field pointing upward (out of the ground). Using the Right Hand Palm Rule, determine the direction of the force on the wire. Then explain what would happen if: (a) the current direction were reversed, (b) the magnetic field direction were reversed.

Show Answer

Initial situation:

Fingers: Point upward (direction of B-field)

Thumb: Point east (direction of current)

Palm: Pushes toward the south

Answer: The force on the wire is directed southward.

(a) Current reversed (now west):

Thumb now points west, fingers still point up.

Palm now pushes toward the north.

Result: Force reverses to northward direction.

(b) Magnetic field reversed (now downward):

Fingers now point down, thumb points east (original current).

Palm now pushes toward the north.

Result: Force also reverses to northward direction.

Key Insight: Reversing either current OR field reverses the force. Reversing both would keep the force in the original direction (south).

Pillar 2 of 3

The Motor Effect & DC Motors

Harness magnetic forces to create rotational motion through torque on current-carrying coils and understand the essential components of DC motors.

Torque on a Coil

Torque is the rotational equivalent of force. When a current-carrying coil is placed in a magnetic field, it experiences a torque that causes rotation. This is the fundamental principle behind electric motors.

Torque Formula

Definition of Torque:

\tau = Fd_{\perp}

Where:

$\tau$ = torque (N·m)
$F$ = force (N)
$d_{\perp}$ = perpendicular distance from pivot to line of force (m)

Torque on a Current-Carrying Coil:

\tau = nIAB\sin\theta

Where:

$\tau$ = torque (N·m)
$n$ = number of turns in the coil
$I$ = current (A)
$A$ = area of the coil (m²)
$B$ = magnetic field strength (T)
$\theta$ = angle between the normal to the coil and the magnetic field

Critical: Angle Definition

$\theta$ is the angle between the normal (perpendicular) to the coil plane and the magnetic field direction, NOT the angle between the coil plane and the field.

If given the angle between the coil plane and the field (call it $\phi$ ), then $\theta = 90° - \phi$ , so use $\tau = nIAB\cos\phi$ .

Understanding Torque on a Coil:

Maximum Torque

Torque is maximum when $\theta = 90°$ (coil plane is parallel to the field). At this position, $\sin\theta = 1$ , so $\tau_{\max} = nIAB$ .

Zero Torque

Torque is zero when $\theta = 0°$ or $180°$ (coil plane is perpendicular to the field). At these positions, forces on opposite sides of the coil are collinear and cancel rotationally.

Force on Each Side

The force on each current-carrying side of the coil is $F = BIL$ . For a rectangular coil, the two sides perpendicular to the field contribute to torque.

Multiple Turns

Each turn of the coil contributes equally to the total torque. Increasing the number of turns $n$ proportionally increases the torque.

Derivation of Torque Formula

Step 1: Force on each side

Consider a rectangular coil with sides of length $l$ (perpendicular to B) and width $w$ :

Force on each vertical side: $F = BIl$

Step 2: Calculate perpendicular distance

The perpendicular distance from the axis of rotation to each force is:

$d_{\perp} = \frac{w}{2}\sin\theta$

Step 3: Calculate torque from one side

$\tau_1 = F \times d_{\perp} = BIl \times \frac{w}{2}\sin\theta$

Step 4: Total torque from both sides

Both vertical sides contribute equally, so multiply by 2:

$\tau = 2 \times BIl \times \frac{w}{2}\sin\theta = BIlw\sin\theta$

Step 5: Express in terms of area

Since $A = lw$ (area of coil):

\tau = IAB\sin\theta

Step 6: Include multiple turns

For $n$ turns, each contributes the same torque:

\tau = nIAB\sin\theta

Worked Example

A rectangular coil has 50 turns, dimensions 8.0 cm × 6.0 cm, and carries a current of 2.5 A. It is placed in a uniform magnetic field of strength 0.40 T. Calculate the torque when: (a) the normal to the coil makes an angle of 30° with the field, (b) the coil plane makes an angle of 30° with the field.

Given:

$n = 50, \quad A = 0.08 \times 0.06 = 4.8 \times 10^{-3} \text{ m}^2$

$I = 2.5 \text{ A}, \quad B = 0.40 \text{ T}$

(a) Normal makes 30° with field:

Here $\theta = 30°$ directly:

$\tau = nIAB\sin\theta$

$\tau = (50)(2.5)(4.8 \times 10^{-3})(0.40)\sin(30°)$

$\tau = (50)(2.5)(4.8 \times 10^{-3})(0.40)(0.5)$

$\tau = \mathbf{0.12 \text{ N}\cdot\text{m}}$

(b) Coil plane makes 30° with field:

If the plane makes 30° with B, then the normal makes $90° - 30° = 60°$ with B:

$\theta = 60°$

$\tau = (50)(2.5)(4.8 \times 10^{-3})(0.40)\sin(60°)$

$\tau = (50)(2.5)(4.8 \times 10^{-3})(0.40)(0.866)$

$\tau = \mathbf{0.208 \text{ N}\cdot\text{m}}$

Note: Alternatively for part (b), you could use $\tau = nIAB\cos(30°)$ directly.

Practice Question

A circular coil of radius 5.0 cm with 100 turns carries a current of 1.5 A in a magnetic field of 0.60 T. At what angle (between the normal and the field) does the coil experience a torque of 0.35 N·m?

Show Answer

Solution:

Area of circular coil: $A = \pi r^2 = \pi (0.05)^2 = 7.85 \times 10^{-3} \text{ m}^2$

Using $\tau = nIAB\sin\theta$ :

$0.35 = (100)(1.5)(7.85 \times 10^{-3})(0.60)\sin\theta$

$0.35 = 0.7065\sin\theta$

$\sin\theta = \frac{0.35}{0.7065} = 0.495$

$\theta = \arcsin(0.495) = \mathbf{29.7°}$

DC Motor Anatomy

A DC motor converts electrical energy into mechanical rotational energy using the motor effect. Understanding each component's function is essential for explaining how motors work and troubleshooting motor behavior.

Essential Components:

Stator (Field Magnets)

Function: Provides the external magnetic field in which the coil rotates.

Types: Can be permanent magnets (simple motors) or electromagnets (stronger, controllable field). The stator is stationary (hence the name).

Armature/Rotor

Function: The rotating coil(s) that carry current and experience the motor force.

Construction: Usually wound around an iron core to increase magnetic field strength and provide structural support. Mounted on a shaft that delivers mechanical power.

Split-Ring Commutator

Function: Reverses the current direction in the coil every half rotation to maintain unidirectional torque.

Why needed: Without it, the coil would oscillate back and forth. The commutator ensures the coil always experiences torque in the same rotational direction.

Brushes

Function: Provide sliding electrical contact between the stationary external circuit and the rotating commutator.

Material: Usually made of carbon (graphite) which is conductive, self-lubricating, and wears slowly. Brushes eventually need replacement due to friction.

How a DC Motor Works

Step 1: Initial Position (Maximum Torque)

When the coil plane is parallel to the magnetic field (normal perpendicular to B), current flows through the coil creating maximum torque. The forces on opposite sides of the coil create a couple that rotates the coil.

Step 2: Rotation Continues

As the coil rotates, the angle changes and torque decreases following $\tau = nIAB\sin\theta$ . The coil's momentum carries it through the position of zero torque (when coil is perpendicular to B).

Step 3: Commutator Action

At the zero-torque position, the split-ring commutator reverses the current direction. This ensures that when the coil moves past this position, the torque continues in the same rotational direction rather than reversing.

Step 4: Continuous Rotation

The cycle repeats: as the coil approaches perpendicular to the field, current reverses, maintaining torque in the same direction. The motor achieves continuous rotation.

Key Insight: The commutator's timing is critical. It must reverse the current exactly when the coil passes through the zero-torque position to maintain efficient rotation.

Design Improvements:

Multiple Coils

Real motors use multiple coils arranged at different angles around the armature. This provides more consistent torque and smoother operation compared to a single coil.

Iron Core

The armature coil is wound around a soft iron core. This concentrates the magnetic field, increasing the force on the coil and therefore the torque produced.

Curved Pole Pieces

Using curved magnets creates a radial field that keeps the field perpendicular to the coil sides over a wider range of rotation, maintaining higher torque.

Brush Maintenance

Brushes wear down over time due to friction and sparking. Regular inspection and replacement is necessary for motor longevity. Brushless motors eliminate this maintenance issue.

Practice Question

Explain why a DC motor would not work properly if the split-ring commutator was replaced with slip rings (like those in an AC generator). What would happen to the coil's motion?

Show Answer

Answer:

Slip rings maintain continuous electrical contact without reversing current direction. Here's what would happen:

Initial rotation: The coil would start rotating due to the motor effect, experiencing torque that rotates it toward the perpendicular position.
Past perpendicular: Once the coil passes the perpendicular position (zero torque), the current direction remains unchanged (unlike with a commutator).
Torque reversal: With the same current but reversed geometry, the torque now acts in the opposite rotational direction, slowing the coil down.
Oscillation: The coil would oscillate back and forth around the perpendicular position rather than rotating continuously.

Conclusion: The split-ring commutator is essential for unidirectional rotation. By reversing current every half-turn, it ensures the torque always acts in the same rotational direction, enabling continuous rotation rather than oscillation.

Radial Magnetic Fields

In a radial magnetic field, the field lines point radially inward or outward from a central axis. This geometry ensures that the magnetic field is always perpendicular to the coil sides, maximizing torque throughout rotation.

Creating a Radial Field

Method: Curved Pole Pieces

Instead of flat magnets, use curved (cylindrical) pole pieces that wrap around the rotating coil. The magnetic field lines emerge radially from the cylindrical poles.

Cylindrical Iron Core

A soft iron cylindrical core is placed at the center. The field lines pass through this core and emerge radially, creating the desired field geometry.

Result: Constant Perpendicular Orientation

As the coil rotates, the sides always cut through field lines perpendicularly. This means $\theta = 90°$ is maintained, so $\sin\theta = 1$ throughout the rotation.

Advantages of Radial Fields:

Maximum Torque

Since $\theta = 90°$ is maintained, the torque is always at its maximum value: $\tau = nIAB$ (without the $\sin\theta$ factor reducing it).

Smooth Operation

Constant torque eliminates the pulsing effect that occurs in simple motors where torque varies with position. This provides smoother, more efficient operation.

Wider Operating Range

The perpendicular geometry is maintained over a much wider range of rotation angles, not just at specific positions as in uniform fields.

Galvanometer Application

Galvanometers (sensitive current meters) use radial fields to ensure torque is proportional to current regardless of pointer position, enabling accurate linear scales.

Uniform Field vs Radial Field

Property	Uniform Field	Radial Field
Field Geometry	Parallel field lines	Radial field lines
Torque Variation	$\tau = nIAB\sin\theta$ (varies with position)	$\tau = nIAB$ (constant)
Max Torque Position	Only when $\theta = 90°$	All positions
Zero Torque Position	When $\theta = 0°, 180°$	Never (always perpendicular)
Application	Simple motors, demonstrations	Galvanometers, precision motors

Practice Question

A galvanometer uses a coil with 200 turns, area 1.5 cm², in a radial magnetic field of 0.25 T. When a current of 50 μA flows through the coil, what torque does it experience? Why is a radial field essential for galvanometer accuracy?

Show Answer

Torque calculation:

Convert area: $A = 1.5 \text{ cm}^2 = 1.5 \times 10^{-4} \text{ m}^2$

Convert current: $I = 50 \text{ }\mu\text{A} = 50 \times 10^{-6} \text{ A}$

In a radial field, $\theta = 90°$ always:

$\tau = nIAB = (200)(50 \times 10^{-6})(1.5 \times 10^{-4})(0.25)$

$\tau = \mathbf{3.75 \times 10^{-7} \text{ N}\cdot\text{m}}$

Why radial fields are essential:

Linear relationship: The torque is directly proportional to current at all positions: $\tau \propto I$ . This allows the galvanometer scale to be linear and uniform.

Position independence: In a uniform field, torque varies with coil position ( $\sin\theta$ ). The reading would depend on where the pointer is, making accurate measurements impossible.

Conclusion: The radial field ensures that torque (and hence deflection) depends only on the current being measured, not on the coil's angular position, enabling accurate, reliable current measurement.

Back EMF in Motors

When a motor runs, the rotating coil cuts through magnetic field lines, inducing an EMF by Faraday's Law. This back EMF (or counter-EMF) opposes the applied voltage and significantly affects motor current and performance.

Understanding Back EMF

Origin of Back EMF:

As the motor coil rotates in the magnetic field, it experiences a changing magnetic flux. By Faraday's Law, this induces an EMF. By Lenz's Law, this induced EMF opposes the change causing it—opposing the applied voltage.

Effective Voltage:

V_{\text{net}} = V_{\text{applied}} - V_{\text{back}}

The current through the motor depends on this net voltage:

I = \frac{V_{\text{net}}}{R} = \frac{V_{\text{applied}} - V_{\text{back}}}{R}

Back EMF Magnitude:

Back EMF is proportional to the motor's rotational speed. Faster rotation means more rapid flux change, inducing larger back EMF: $V_{\text{back}} \propto \omega$

Motor Behavior at Different Speeds:

Start-Up (ω = 0)

Back EMF: Zero (no rotation, no flux change)

Current: Maximum ( $I = V/R$ )

Risk: High current can overheat windings and damage the motor. This is why motors draw much more current at start-up than during normal operation.

Running (Normal Speed)

Back EMF: Large (approaching $V_{\text{applied}}$ )

Current: Reduced to a safe operating level

Result: Motor runs efficiently with manageable current. The back EMF acts as a natural current limiter.

No Load (Maximum Speed)

Back EMF: Nearly equals $V_{\text{applied}}$

Current: Very small (only enough to overcome friction)

Behavior: Motor reaches maximum speed when back EMF almost balances applied voltage. Any further speed increase would make back EMF exceed applied voltage, reducing current and torque.

Under Load (Reduced Speed)

Back EMF: Decreases (lower speed)

Current: Increases automatically

Self-Regulation: When load increases and motor slows, back EMF decreases, allowing more current to flow, which increases torque to handle the load. This is automatic speed/torque regulation.

Protecting Motors from Start-Up Current

Problem: At start-up, $V_{\text{back}} = 0$ , so $I = V/R$ can be 5-10 times the normal operating current, risking motor burnout.

Solution 1: Starting Resistor

Add a resistor in series during start-up to limit current. Once motor reaches operating speed (and back EMF builds up), remove or bypass the resistor for normal operation.

Solution 2: Reduced Voltage Start

Apply reduced voltage initially, then gradually increase to full voltage as motor speeds up. Modern motor controllers use electronic circuits to implement this smoothly.

Solution 3: Soft-Start Circuits

Electronic circuits (using thyristors or transistors) gradually ramp up the voltage over a second or two, allowing back EMF to build naturally and limiting current surge.

Worked Example

A DC motor has a coil resistance of 5.0 Ω and is connected to a 120 V supply. Calculate the current drawn: (a) at start-up when the motor is stationary, (b) when running at normal speed with a back EMF of 110 V, (c) the percentage reduction in current.

Given:

$R = 5.0 \text{ }\Omega, \quad V_{\text{applied}} = 120 \text{ V}, \quad V_{\text{back}} = 110 \text{ V (running)}$

(a) Start-up current:

At start-up, $V_{\text{back}} = 0$ :

$I_{\text{start}} = \frac{V_{\text{applied}}}{R} = \frac{120}{5.0}$

$I_{\text{start}} = \mathbf{24 \text{ A}}$

(b) Running current:

$I_{\text{run}} = \frac{V_{\text{applied}} - V_{\text{back}}}{R}$

$I_{\text{run}} = \frac{120 - 110}{5.0} = \frac{10}{5.0}$

$I_{\text{run}} = \mathbf{2.0 \text{ A}}$

$\text{Reduction} = \frac{I_{\text{start}} - I_{\text{run}}}{I_{\text{start}}} \times 100\%$

$\text{Reduction} = \frac{24 - 2.0}{24} \times 100\%$

$\text{Reduction} = \frac{22}{24} \times 100\% = \mathbf{91.7\%}$

The current reduces by over 90% once the motor is running! The start-up current (24 A) is 12 times the running current (2 A), demonstrating why start-up protection is essential.

Practice Question

A motor connected to a 240 V supply draws 30 A at start-up. When running normally, it draws 4.0 A. Calculate: (a) the coil resistance, (b) the back EMF when running normally, (c) explain what happens to the motor speed if a mechanical load is added.

Show Answer

(a) Coil resistance:

At start-up, $V_{\text{back}} = 0$ :

$R = \frac{V}{I_{\text{start}}} = \frac{240}{30} = \mathbf{8.0 \text{ }\Omega}$

(b) Back EMF when running:

$I_{\text{run}} = \frac{V_{\text{applied}} - V_{\text{back}}}{R}$

$4.0 = \frac{240 - V_{\text{back}}}{8.0}$

$32 = 240 - V_{\text{back}}$

$V_{\text{back}} = \mathbf{208 \text{ V}}$

When mechanical load is added:

The motor slows down (increased resistance to rotation)
Back EMF decreases (proportional to speed)
Net voltage increases: $V_{\text{net}} = V_{\text{applied}} - V_{\text{back}} \uparrow$
Current increases: $I = V_{\text{net}}/R \uparrow$
Torque increases: $\tau = nIAB \uparrow$
Motor finds new equilibrium at lower speed but higher torque to handle the load

Self-regulation: The motor automatically adjusts current and torque to match the load requirement, slowing down slightly under heavier loads.

Pillar 3 of 3

Electromagnetic Induction

Convert mechanical energy to electrical energy through Faraday's Law, understand transformers, and explore AC/DC generators and induction motors.

Magnetic Flux

Magnetic flux (symbol $\Phi$ ) is a measure of the total magnetic field passing through a given area. It quantifies how much magnetic field "flows through" a surface and is fundamental to understanding electromagnetic induction.

Magnetic Flux Formula

\Phi = BA\cos\theta

Where:

$\Phi$ = magnetic flux (Wb, Weber)
$B$ = magnetic field strength (T, Tesla)
$A$ = area (m²)
$\theta$ = angle between the normal to the area and the magnetic field direction

Critical: Angle Definition

$\theta$ is the angle between the normal (perpendicular) to the surface and the magnetic field $B$ , NOT the angle between the surface itself and the field.

Maximum flux occurs when $\theta = 0°$ (normal parallel to $B$ , surface perpendicular to $B$ ): $\Phi_{\max} = BA$

Zero flux occurs when $\theta = 90°$ (normal perpendicular to $B$ , surface parallel to $B$ ): $\Phi = 0$

Unit: 1 Weber (Wb) = 1 T·m² = 1 kg·m²/(A·s²)

Understanding Magnetic Flux:

Physical Interpretation

Flux represents the "amount" of magnetic field passing through an area. Imagine field lines as flowing water—flux measures how much flows through a given surface.

Maximum Flux Position

Flux is maximum when the surface is perpendicular to the field lines (normal parallel to $B$ ). This is when the most field lines pass through the area.

Changing Flux

Flux can change by: (1) changing $B$ , (2) changing $A$ , or (3) changing $\theta$ (rotating the coil). All three methods induce EMF.

Sign Convention

The direction of the normal vector determines the sign of flux. By convention, choose a consistent direction for the normal when analyzing a problem.

Worked Example

A rectangular coil with dimensions 15 cm × 10 cm is placed in a uniform magnetic field of 0.60 T. Calculate the magnetic flux through the coil when: (a) the normal to the coil is parallel to the field, (b) the normal makes an angle of 30° with the field, (c) the coil plane is parallel to the field.

Given:

$A = 0.15 \times 0.10 = 0.015 \text{ m}^2, \quad B = 0.60 \text{ T}$

(a) Normal parallel to field ( $\theta = 0°$ ):

$\Phi = BA\cos\theta = BA\cos(0°)$

$\Phi = (0.60)(0.015)(1)$

$\Phi = \mathbf{9.0 \times 10^{-3} \text{ Wb} = 9.0 \text{ mWb}}$

This is maximum flux.

(b) Normal makes 30° with field:

$\Phi = BA\cos\theta = (0.60)(0.015)\cos(30°)$

$\Phi = (0.60)(0.015)(0.866)$

$\Phi = \mathbf{7.79 \times 10^{-3} \text{ Wb} = 7.79 \text{ mWb}}$

If the plane is parallel to $B$ , the normal is perpendicular to $B$ :

$\theta = 90°$

$\Phi = BA\cos(90°) = BA(0) = \mathbf{0 \text{ Wb}}$

No field lines pass through the coil.

Practice Question

A circular coil of radius 8.0 cm has its plane initially perpendicular to a uniform magnetic field of 0.50 T. The coil is then rotated through 60° about an axis in the plane of the coil. Calculate: (a) the initial flux, (b) the final flux, (c) the change in flux.

Show Answer

Setup:

Area: $A = \pi r^2 = \pi (0.08)^2 = 0.0201 \text{ m}^2$

Initially: plane perpendicular to $B$ means normal parallel to $B$ ( $\theta_i = 0°$ )

Finally: rotated 60° means $\theta_f = 60°$

(a) Initial flux:

$\Phi_i = BA\cos(0°) = (0.50)(0.0201)(1)$

$\Phi_i = \mathbf{0.0101 \text{ Wb} = 10.1 \text{ mWb}}$

(b) Final flux:

$\Phi_f = BA\cos(60°) = (0.50)(0.0201)(0.5)$

$\Phi_f = \mathbf{0.00503 \text{ Wb} = 5.03 \text{ mWb}}$

$\Delta\Phi = \Phi_f - \Phi_i = 5.03 - 10.1$

$\Delta\Phi = \mathbf{-5.07 \text{ mWb}}$

The negative sign indicates flux has decreased.

Faraday's Law

Faraday's Law of Electromagnetic Induction states that a changing magnetic flux through a coil induces an electromotive force (EMF) in the coil. This is the fundamental principle behind generators and transformers.

Faraday's Law Formula

\varepsilon = -N \frac{\Delta\Phi}{\Delta t}

Where:

$\varepsilon$ = induced EMF (V)
$N$ = number of turns in the coil
$\Delta\Phi$ = change in magnetic flux (Wb)
$\Delta t$ = time interval (s)
Negative sign represents Lenz's Law (discussed next section)

Key Insight: EMF is induced only when flux is changing. A constant flux, no matter how large, produces zero EMF. The rate of change matters.

Methods of Changing Flux:

Change Magnetic Field (B)

Increase or decrease the magnetic field strength through the coil. Example: Moving a magnet toward or away from a coil changes $B$ , inducing EMF.

Change Area (A)

Change the effective area of the coil in the field. Example: A sliding contact that changes the area of a circuit loop in a magnetic field.

Change Angle (θ)

Rotate the coil in the magnetic field. Example: Generators rotate coils through magnetic fields, continuously changing $\theta$ and thus $\Phi = BA\cos\theta$ .

Multiple Turns

Each turn of the coil contributes to the total induced EMF. A coil with $N$ turns produces $N$ times the EMF of a single-turn coil.

Worked Example

A coil with 200 turns has an area of 0.050 m² and is placed perpendicular to a magnetic field. The field strength decreases uniformly from 0.80 T to 0.20 T in 0.30 seconds. Calculate: (a) the change in flux through one turn, (b) the magnitude of the induced EMF.

Given:

$N = 200, \quad A = 0.050 \text{ m}^2, \quad B_i = 0.80 \text{ T}, \quad B_f = 0.20 \text{ T}$

$\Delta t = 0.30 \text{ s}, \quad \theta = 0° \text{ (perpendicular)}$

(a) Change in flux through one turn:

Initial flux: $\Phi_i = B_i A\cos(0°) = (0.80)(0.050)(1) = 0.040 \text{ Wb}$

Final flux: $\Phi_f = B_f A\cos(0°) = (0.20)(0.050)(1) = 0.010 \text{ Wb}$

Change in flux: $\Delta\Phi = \Phi_f - \Phi_i = 0.010 - 0.040$

$\Delta\Phi = \mathbf{-0.030 \text{ Wb}}$

Negative indicates flux is decreasing.

(b) Magnitude of induced EMF:

$|\varepsilon| = N \frac{|\Delta\Phi|}{\Delta t}$

$|\varepsilon| = 200 \times \frac{0.030}{0.30}$

$|\varepsilon| = 200 \times 0.10$

$|\varepsilon| = \mathbf{20 \text{ V}}$

Practice Question

A single-turn circular coil of radius 10 cm is rotated from a position perpendicular to a 0.40 T magnetic field to a position parallel to the field in 0.15 seconds. Calculate the average induced EMF during this rotation.

Show Answer

Solution:

Area: $A = \pi r^2 = \pi (0.10)^2 = 0.0314 \text{ m}^2$

Initial position (perpendicular to $B$ ): $\theta_i = 0°$

$\Phi_i = BA\cos(0°) = (0.40)(0.0314)(1) = 0.01257 \text{ Wb}$

Final position (parallel to $B$ ): $\theta_f = 90°$

$\Phi_f = BA\cos(90°) = 0 \text{ Wb}$

Change in flux:

$\Delta\Phi = 0 - 0.01257 = -0.01257 \text{ Wb}$

Induced EMF (N = 1 turn):

$|\varepsilon| = \frac{|\Delta\Phi|}{\Delta t} = \frac{0.01257}{0.15}$

$|\varepsilon| = \mathbf{0.0838 \text{ V} = 83.8 \text{ mV}}$

Lenz's Law

Lenz's Law explains the negative sign in Faraday's Law and states that the induced EMF (and resulting current) always acts to oppose the change in magnetic flux that produced it. This is a consequence of energy conservation.

Understanding Lenz's Law

The Negative Sign in Faraday's Law:

\varepsilon = -N \frac{\Delta\Phi}{\Delta t}

The negative sign indicates that the induced EMF creates a current whose magnetic field opposes the flux change causing it.

Two Scenarios:

Flux Increasing:

The induced current creates a magnetic field that opposes the increase—pointing opposite to the external field.

Flux Decreasing:

The induced current creates a magnetic field that opposes the decrease—pointing in the same direction as the external field (trying to maintain it).

Energy Conservation: If the induced current aided the flux change instead of opposing it, we'd have a runaway effect creating energy from nothing, violating conservation of energy.

Step-by-Step Lenz's Law Analysis

To determine the direction of induced current:

Step 1: Identify the change in flux

Is the magnetic flux through the coil increasing or decreasing? In what direction is the external field pointing?

Step 2: Determine the opposing field

If flux is increasing, the induced field must point opposite to the external field. If flux is decreasing, the induced field must point in the same direction as the external field.

Step 3: Find current direction

Use the right-hand grip rule: curl your fingers in the direction of current flow, and your thumb points in the direction of the magnetic field created by that current. Work backwards to find the current direction that produces the required opposing field.

Example Template for Exam Answers:

"The magnetic flux through the coil is [increasing/decreasing] in the [upward/downward/...] direction. By Lenz's Law, the induced current must create a magnetic field that [opposes this increase/opposes this decrease], pointing [direction]. Using the right-hand grip rule, this requires a current flowing [clockwise/counterclockwise/...]."

Common Applications:

Magnet Approaching Coil

As north pole approaches, flux increases upward. Induced current creates downward field (opposing increase). Current flows to create a north pole facing the approaching magnet, repelling it.

Magnet Leaving Coil

As north pole moves away, flux decreases. Induced current creates field in same direction as magnet (opposing decrease). Current flows to create a south pole facing the departing magnet, attracting it.

Eddy Currents

When a conductor moves through a magnetic field, circular eddy currents are induced that create forces opposing the motion. This provides electromagnetic braking.

Back EMF in Motors

The rotating coil in a motor induces a back EMF that opposes the applied voltage. This is Lenz's Law ensuring the motor doesn't accelerate infinitely.

Worked Example

A bar magnet is dropped vertically downward through a horizontal coil with its north pole pointing down. Describe the direction of the induced current (viewed from above) as: (a) the north pole approaches the coil, (b) the north pole passes through the coil, (c) the north pole moves away below the coil.

(a) North pole approaching from above:

Flux change: Downward magnetic flux through coil is increasing.

Lenz's Law: Induced current must create an upward magnetic field to oppose this increase.

Right-hand grip rule: For upward field, current must flow counterclockwise (viewed from above).

Effect: The coil's induced field creates a north pole on top, repelling the approaching north pole of the magnet.

Answer: Counterclockwise (viewed from above)

(b) North pole passing through coil:

At the exact moment of passing through, the flux is momentarily constant (not increasing or decreasing), so no current is induced at this instant.

Flux change: Downward magnetic flux through coil is decreasing.

Lenz's Law: Induced current must create a downward magnetic field to oppose this decrease (maintain flux).

Right-hand grip rule: For downward field, current must flow clockwise (viewed from above).

Effect: The coil's induced field creates a south pole on top, attracting the departing north pole of the magnet.

Answer: Clockwise (viewed from above)

Key Insight: The induced current direction reverses as the magnet passes through. The coil always acts to oppose the magnet's motion—repelling when approaching, attracting when leaving.

Practice Question

A metal ring is placed on a table. A bar magnet is held above the ring with its south pole pointing down and is then suddenly dropped toward the ring. Using Lenz's Law, explain: (a) the direction of current induced in the ring, (b) whether the ring is attracted to or repelled by the falling magnet, (c) what happens to the magnet's acceleration compared to free fall.

Show Answer

(a) Direction of induced current:

As the south pole approaches, upward magnetic flux through the ring increases (south pole below means field lines point upward through the ring).

By Lenz's Law, the induced current creates a downward magnetic field to oppose this increase.

Using the right-hand grip rule, the current flows clockwise when viewed from above.

(b) Attraction or repulsion:

The clockwise current creates a south pole on the top surface of the ring.

Since the approaching magnet has its south pole pointing down, there is repulsion between like poles.

The ring is repelled by the magnet (or equivalently, the magnet is repelled upward by the ring).

The magnetic repulsion creates an upward force on the falling magnet.

This opposes gravity, so the net downward force is less than $mg$ .

Therefore, the magnet's acceleration is less than g (free fall)—it falls more slowly than it would without the ring.

Energy perspective: The magnet's gravitational potential energy converts partly to kinetic energy and partly to electrical energy (current in ring) and heat. Less energy goes to kinetic energy, so it accelerates more slowly.

Ideal Transformers

A transformer uses electromagnetic induction to change AC voltage levels. An ideal transformer has no energy losses, making it 100% efficient. Transformers only work with AC (alternating current) because they require changing magnetic flux.

Transformer Equations

Voltage-Turns Relationship:

\frac{V_p}{V_s} = \frac{N_p}{N_s}

Where:

$V_p$ = primary voltage (input)
$V_s$ = secondary voltage (output)
$N_p$ = number of turns in primary coil
$N_s$ = number of turns in secondary coil

Current-Turns Relationship (Ideal Transformer):

\frac{I_s}{I_p} = \frac{N_p}{N_s}

Note: Current ratio is inverse of voltage ratio!

Power Conservation (Ideal Transformer):

P_p = P_s

V_p I_p = V_s I_s

In an ideal transformer, power in equals power out (100% efficiency).

Critical Exam Habit:

Always state "Assuming an ideal transformer (100% efficiency)" before using $V_p I_p = V_s I_s$ . This shows the examiner you understand real transformers have losses.

Transformer Types:

Step-Up Transformer

Purpose: Increase voltage

Design: $N_s > N_p$ (more turns in secondary)

Effect: $V_s > V_p$ but $I_s < I_p$

Example: Power stations step up voltage to 500,000V for transmission

Step-Down Transformer

Purpose: Decrease voltage

Design: $N_s < N_p$ (fewer turns in secondary)

Effect: $V_s < V_p$ but $I_s > I_p$

Example: Household transformers step down from 240V to 12V for devices

How Transformers Work

Step 1: AC voltage applied to primary coil creates alternating current

Step 2: Alternating current in primary creates changing magnetic field in iron core

Step 3: Changing magnetic flux passes through secondary coil (via iron core)

Step 4: By Faraday's Law, changing flux induces EMF in secondary coil

Step 5: Induced EMF drives current in secondary circuit (if connected to load)

Why AC only?

DC produces constant magnetic field (no flux change), so no EMF is induced in secondary. Transformers only work with AC because they require $\Delta\Phi/\Delta t \neq 0$ .

Worked Example

An ideal transformer has 200 turns in the primary coil and 50 turns in the secondary coil. The primary is connected to 240 V AC and draws a current of 2.0 A. Calculate: (a) the secondary voltage, (b) the secondary current, (c) the power in the primary and secondary.

Given:

$N_p = 200, \quad N_s = 50, \quad V_p = 240 \text{ V}, \quad I_p = 2.0 \text{ A}$

(a) Secondary voltage:

$\frac{V_p}{V_s} = \frac{N_p}{N_s}$

$\frac{240}{V_s} = \frac{200}{50}$

$V_s = \frac{240 \times 50}{200}$

$V_s = \mathbf{60 \text{ V}}$

This is a step-down transformer (voltage decreased by factor of 4).

(b) Secondary current:

Assuming an ideal transformer (100% efficiency):

$\frac{I_s}{I_p} = \frac{N_p}{N_s}$

$\frac{I_s}{2.0} = \frac{200}{50}$

$I_s = 2.0 \times 4$

$I_s = \mathbf{8.0 \text{ A}}$

Current increased by factor of 4 (inverse of voltage ratio).

Primary: $P_p = V_p I_p = (240)(2.0) = 480 \text{ W}$

Secondary: $P_s = V_s I_s = (60)(8.0) = 480 \text{ W}$

$P_p = P_s = \mathbf{480 \text{ W}}$

Power is conserved (100% efficiency in ideal transformer).

Practice Question

A transformer is needed to step up 12 V DC from a car battery to 240 V for household appliances. The secondary coil has 2000 turns. (a) How many turns should the primary have? (b) Will this transformer work? Explain.

Show Answer

(a) Number of primary turns:

Using $\frac{V_p}{V_s} = \frac{N_p}{N_s}$ :

$\frac{12}{240} = \frac{N_p}{2000}$

$N_p = \frac{12 \times 2000}{240}$

$N_p = \mathbf{100 \text{ turns}}$

(b) Will it work?

NO, this transformer will NOT work.

Reason: Transformers only work with AC (alternating current), not DC (direct current).

DC produces a constant magnetic field in the iron core. Since the flux is not changing ( $\Delta\Phi/\Delta t = 0$ ), Faraday's Law shows no EMF is induced in the secondary coil ( $\varepsilon = -N\Delta\Phi/\Delta t = 0$ ).

Solution: First convert the 12V DC to 12V AC using an inverter, then use the transformer.

Real Transformers: Energy Losses

Real transformers are not 100% efficient. Various energy losses reduce efficiency to typically 95-99%. Understanding these losses and their solutions is essential for designing efficient power distribution systems.

Three Main Energy Losses:

1. Flux Leakage

Problem: Not all magnetic flux from the primary passes through the secondary coil. Some flux "leaks" into the surrounding air instead of following the core path.

Effect: Reduced induced EMF in secondary (less than ideal value). Lower efficiency and voltage output.

Solution: Iron Core

Use a soft iron core to provide a low-reluctance (high permeability) path for magnetic flux. Iron concentrates and guides the field lines, ensuring nearly all flux from primary reaches secondary. Typical efficiency improvement: 90% → 98%.

2. Resistive Heating (I²R Losses)

Problem: Copper wire in coils has resistance. Current flowing through produces heat: $P = I^2R$ . This energy is wasted as thermal energy.

Effect: Energy converted to heat in coils instead of being transferred to secondary. Coils can overheat under heavy load.

Solution: Thick Copper Wires

Use thick (low resistance) copper wires for windings. Since $R \propto 1/A$ (inversely proportional to cross-sectional area), thicker wire means lower resistance and less heat loss: $P_{\text{loss}} = I^2R \downarrow$ .

3. Eddy Current Losses

Problem: The changing magnetic flux induces circular currents (eddy currents) in the iron core itself. These currents flow through the core's resistance, producing heat: $P = I^2R$ .

Effect: Significant energy loss as heat in the core. Can cause dangerous overheating in solid iron cores.

Solution: Laminated Iron Core

Build the core from thin sheets (laminations) of iron separated by insulating layers (varnish/oxide). This breaks up the eddy current paths, forcing currents to flow in much smaller loops. Since eddy current magnitude is proportional to loop area, smaller loops mean much smaller currents and dramatically reduced losses.

Laminations are typically 0.3-0.5 mm thick, reducing eddy current losses by 95%+.

Additional Energy Losses

Hysteresis Losses:

Energy is required to repeatedly magnetize and demagnetize the iron core as the AC current alternates. The iron domains resist alignment changes, dissipating energy as heat.

Solution: Use soft iron or silicon steel which magnetizes/demagnetizes easily (low coercivity).

Sound/Vibration:

The alternating magnetic field causes the core laminations to vibrate, producing the characteristic "hum" of transformers and dissipating small amounts of energy as sound.

Solution: Tight mechanical assembly, vibration damping materials.

Ideal vs Real Transformers

Property	Ideal Transformer	Real Transformer
Efficiency	100%	95-99%
Flux Linkage	Perfect (all flux links)	Some flux leakage
Coil Resistance	Zero	Finite (causes I²R losses)
Core Losses	None	Eddy currents, hysteresis
Power Relation	$P_p = P_s$	$P_p > P_s$

Practice Question

A power transformer has a primary drawing 10,000 W and delivers 9,500 W to the secondary. (a) Calculate the efficiency. (b) Identify three specific design features that would improve this efficiency and explain how each one works.

Show Answer

(a) Efficiency:

$\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\%$

$\text{Efficiency} = \frac{9500}{10000} \times 100\%$

$\text{Efficiency} = \mathbf{95\%}$

500 W is lost to heat and other losses.

(b) Three improvements:

1. Laminated iron core:

Use thin insulated iron sheets instead of solid core. This breaks up eddy current paths, reducing circular currents induced in the core. Smaller eddy currents mean less I²R heating in the core, reducing energy loss.

2. Thicker copper wire:

Use thick, low-resistance copper for windings. Lower resistance means less I²R heating in the coils. For example, doubling wire diameter reduces resistance by factor of 4, cutting resistive losses to 1/4.

3. Soft iron or silicon steel core:

Use magnetically soft materials that magnetize and demagnetize easily. This reduces hysteresis losses (energy wasted in repeatedly flipping magnetic domains). Silicon steel can reduce hysteresis losses by 50%.

Generators: AC vs DC

Generators convert mechanical energy (rotation) into electrical energy using electromagnetic induction. As a coil rotates in a magnetic field, the changing flux induces an EMF. The key difference between AC and DC generators is how they deliver current to the external circuit.

Generator Principle

Common Elements (Both AC and DC):

Rotating coil (armature) in a magnetic field
As coil rotates, angle $\theta$ changes continuously
Flux through coil: $\Phi = BA\cos\theta = BA\cos(\omega t)$
By Faraday's Law: $\varepsilon = -N\frac{d\Phi}{dt}$
This produces sinusoidal EMF: $\varepsilon = NBA\omega\sin(\omega t)$

Maximum EMF:

\varepsilon_{\max} = NBA\omega

Where $\omega = 2\pi f$ is the angular velocity of rotation.

AC Generator (Alternator)

Key Component: Slip Rings

Construction: Two complete conducting rings attached to the rotating shaft. Each end of the coil connects to its own slip ring. Carbon brushes maintain continuous contact with both rings.

Function: Slip rings maintain continuous electrical connection while allowing the coil to rotate freely. Each ring is always connected to the same end of the coil.

Result: The output alternates as the coil rotates—positive half-cycle when one end is higher potential, negative half-cycle when reversed.

Output Characteristics:

Sinusoidal AC voltage: $V = V_{\max}\sin(\omega t)$
Frequency matches rotation frequency: $f = \frac{\omega}{2\pi}$
Used in power stations (50 Hz or 60 Hz AC)
Can be stepped up/down with transformers

DC Generator

Key Component: Split-Ring Commutator

Construction: A single ring split into two halves, insulated from each other. Each half connects to one end of the coil. The split aligns with the position where EMF crosses zero.

Function: Every half rotation (when EMF would reverse), the commutator swaps which half connects to which brush. This reverses the external connection at exactly the moment the coil EMF reverses.

Result: The output always has the same polarity—it "flips" the negative half-cycles to positive, producing pulsed DC.

Output Characteristics:

Pulsed DC (unidirectional but varying magnitude)
Voltage varies from zero to maximum, never negative
Can be smoothed with capacitors for more constant DC
Cannot be transformed (transformers need AC)

AC vs DC Generator Comparison

Feature	AC Generator	DC Generator
Contact System	Slip rings (2 complete rings)	Split-ring commutator
Output Type	Sinusoidal AC	Pulsed DC (rectified)
Polarity	Alternates (± voltage)	Constant (always +)
Transformer Compatible	Yes	No
Typical Use	Power stations, grid electricity	Battery charging, small applications
Brush Wear	Less (continuous contact)	More (switching sparks)

Practice Question

A generator coil with 500 turns has an area of 0.025 m² and rotates at 50 revolutions per second in a magnetic field of 0.30 T. Calculate: (a) the maximum EMF produced, (b) the frequency of the AC output if slip rings are used.

Show Answer

(a) Maximum EMF:

Angular velocity: $\omega = 2\pi f = 2\pi(50) = 314.16 \text{ rad/s}$

Using $\varepsilon_{\max} = NBA\omega$ :

$\varepsilon_{\max} = (500)(0.30)(0.025)(314.16)$

$\varepsilon_{\max} = \mathbf{1178 \text{ V}}$

(b) AC frequency:

The frequency of the AC output equals the rotation frequency:

$f = \mathbf{50 \text{ Hz}}$

Each complete rotation produces one complete AC cycle.

AC Induction Motors

AC induction motors are the workhorses of industry, powering everything from refrigerators to factory machinery. Unlike DC motors, they have no brushes or commutators, making them more reliable and maintenance-free. They work through electromagnetic induction rather than direct electrical connection to the rotor.

How AC Induction Motors Work

Step 1: Stator Creates Rotating Magnetic Field

The stator contains three sets of coils arranged 120° apart around the motor. Each coil receives AC current, but the three currents are out of phase (shifted by 120° in time).

As the three AC currents oscillate with different phases, they create a magnetic field that appears to rotate around the motor. This is called a rotating magnetic field.

Step 2: Induction in the Rotor

The rotor (called a "squirrel cage" due to its appearance) consists of conducting bars connected by rings—essentially a set of short-circuited loops.

As the rotating magnetic field sweeps past the rotor bars, the magnetic flux through each loop changes. By Faraday's Law, this changing flux induces EMF in the rotor bars.

The induced EMF drives currents through the short-circuited rotor bars.

Step 3: Motor Effect Produces Torque

The induced currents in the rotor bars are themselves in a magnetic field (the stator's rotating field).

By the motor effect ( $F = BIL$ ), these current-carrying conductors experience forces. By Lenz's Law, these forces act to reduce the relative motion between the rotor and the rotating field.

Result: The rotor is dragged along by the rotating magnetic field, causing it to spin.

Step 4: Slip

The rotor never quite catches up to the rotating field. If it did, there would be no relative motion, no changing flux, and no induced current—thus no torque.

The rotor rotates slightly slower than the rotating field. This speed difference is called slip and is essential for motor operation.

Key Components:

Stator

Structure: Three sets of coils (3-phase windings) arranged symmetrically around a cylindrical iron core.

Function: Creates the rotating magnetic field when supplied with 3-phase AC power. The field rotates at synchronous speed: $n_s = \frac{120f}{p}$ rpm, where $f$ is frequency and $p$ is number of pole pairs.

Squirrel Cage Rotor

Structure: Aluminum or copper bars embedded in an iron rotor core, with conducting rings short-circuiting the bars at both ends. Resembles a hamster wheel—hence "squirrel cage."

Function: Conducts induced currents that interact with the magnetic field to produce torque. No electrical connections needed—current is induced.

No Brushes or Commutator

Unlike DC motors, induction motors have no brushes, slip rings, or commutators. The rotor has no electrical connections at all—it operates purely through induction. This eliminates brush wear and sparking.

Self-Starting

When 3-phase power is applied, the rotating field immediately appears and begins inducing currents. The motor starts automatically without any starter mechanism (for most designs).

Advantages of AC Induction Motors

Robust & Reliable

No brushes or commutator means no wearing parts in electrical contact. Can run for decades with minimal maintenance.

Simple Construction

Squirrel cage rotor is just bars and rings—very simple and cheap to manufacture. No complex windings or electrical connections on rotor.

No Sparking

No brushes means no electrical sparks, making them safe in explosive environments (grain elevators, chemical plants).

Constant Speed

Speed depends on AC frequency and number of poles, remaining nearly constant under varying loads (small slip variation).

Practice Question

Explain the role of Lenz's Law in an AC induction motor. Why does the rotor rotate in the same direction as the rotating magnetic field? Why can't the rotor rotate at the same speed as the rotating field?

Show Answer

Role of Lenz's Law:

Flux change: The rotating magnetic field sweeps past the stationary (or slower-moving) rotor bars, causing the magnetic flux through rotor loops to change continuously.

Induced current: By Faraday's Law, this changing flux induces EMF and current in the rotor bars.

Lenz's Law: The induced current creates its own magnetic field that opposes the change in flux. The way to oppose the relative motion between rotor and field is for the rotor to "chase" the rotating field—to move in the same direction.

Motor effect: The induced current in the rotor bars experiences force in the magnetic field ( $F = BIL$ ), producing torque in the direction of field rotation.

Why rotor can't match field speed:

Suppose the rotor caught up to the rotating field speed. Then:

No relative motion between rotor and field
Flux through rotor loops becomes constant (not changing)
By Faraday's Law: $\Delta\Phi/\Delta t = 0 \implies \varepsilon = 0$
No induced current in rotor bars
No magnetic force on rotor (motor effect requires current)
No torque, so rotor slows down due to friction

Conclusion: The rotor MUST rotate slower than the field to maintain the changing flux that induces current. This speed difference (slip) is essential for operation. Typical slip is 2-5% of synchronous speed.

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Charged Particles, Conductors & Fields

Electric Fields

Key Equations

Direction of Force

Uniform Fields

Field Strength Units

Independence of Charge

Worked Example

Practice Question

Work Done in Electric Fields

Work Done Equations

Accelerating Charge

Decelerating Charge

Sign Convention

The Electron Volt (eV)

Worked Example

Practice Question

Trajectory Analysis in Electric Fields

Motion Analysis

Analogy with Gravity

Deflection Direction

Mass Dependence

CRO Application

Worked Example

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Magnetic Forces on Particles

Magnetic Force on Moving Charges

Force is Perpendicular

Circular Motion

Speed Remains Constant

Charge Sign Matters

Radius of Circular Path

Worked Example

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The Motor Effect (Force on Wires)

Force on a Current-Carrying Conductor

Origin of Force

Direction of Force

Applications

Reversing Force

Worked Example

Practice Question

Force Between Parallel Wires

Force Per Unit Length Between Parallel Wires

Why Parallel Wires Interact

Worked Example

Practice Question

Hand Rules for Electromagnetic Forces

Right Hand Palm Rule (Motor Effect)

Fleming's Left Hand Rule (Alternative)

Right Hand Grip Rule (Magnetic Fields)

Motors

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Parallel Wire Interaction

Solenoids

Practice Question

The Motor Effect & DC Motors

Torque on a Coil

Torque Formula

Maximum Torque

Zero Torque

Force on Each Side

Multiple Turns

Derivation of Torque Formula

Worked Example

Practice Question

DC Motor Anatomy

Stator (Field Magnets)

Armature/Rotor

Split-Ring Commutator

Brushes

How a DC Motor Works

Multiple Coils

Iron Core

Curved Pole Pieces

Brush Maintenance

Practice Question

Radial Magnetic Fields

Creating a Radial Field