Module 4: Electricity & Magnetism

Coulomb's law gives $F \propto \dfrac{1}{r^2}$. Halving $r$ multiplies $\dfrac{1}{r^2}$ by $\dfrac{1}{(1/2)^2} = 4$, so the force becomes $4F$. (The inverse-square law makes small distance changes have a large effect.)

$F = \dfrac{kq_1q_2}{r^2} = \dfrac{9.0\times10^9 \times (2.0\times10^{-6})(3.0\times10^{-6})}{(0.10)^2} = \dfrac{9.0\times10^9 \times 6.0\times10^{-12}}{0.010} = 5.4\text{ N}$. (Watch the $\mu\text{C} = 10^{-6}$ C conversions and the $r^2$.)

Electric field is force per unit charge: $E = \dfrac{F}{q} = \dfrac{0.020}{4.0\times10^{-6}} = 5000\text{ N/C}$. (Dividing by a small number gives a large field; multiplying instead is the common slip.)

Between parallel plates the field is uniform: $E = \dfrac{V}{d} = \dfrac{100}{0.020} = 5000\text{ V/m}$. (The gap must be in metres: $2.0\text{ cm}$ = $0.020\text{ m}$, where a factor of 100 is easily dropped.)

$W = qV = 5.0\times10^{-6} \times 200 = 1.0\times10^{-3}\text{ J}$ = $1.0\text{ mJ}$. (Potential difference is the energy transferred per coulomb, so multiply charge by voltage.)

Each charge produces a field at the midpoint, but the two fields are equal in size and point in opposite directions (each pushing a positive test charge away from itself). They cancel exactly, so the net field is zero. (The electric potential there is not zero, but the field is.)

To float, the electric force balances gravity: $qE = mg$, so $q = \dfrac{mg}{E} = \dfrac{3.3\times10^{-15} \times 9.8}{2.0\times10^4} \approx 1.6\times10^{-18}$ C. The number of electrons is $\dfrac{q}{e} = \dfrac{1.6\times10^{-18}}{1.6\times10^{-19}} = 10$. (This two-step method is the idea behind Millikan's experiment.)

$E = \dfrac{kQ}{r^2} = \dfrac{9.0\times10^9 \times 5.0\times10^{-6}}{(0.30)^2} = \dfrac{45000}{0.09} = 5.0\times10^5\text{ N/C}$. (Field from a point charge follows the inverse-square law, unlike the uniform field between plates.)

Ohm's law: $I = \dfrac{V}{R} = \dfrac{12}{4.0} = 3.0\text{ A}$.

In series the total resistance is $4.0 + 8.0 = 12\ \Omega$, so the current is $I = \dfrac{12}{12} = 1.0\text{ A}$ (the same everywhere). The voltage across the $8.0\ \Omega$ resistor is $V = IR = 1.0 \times 8.0 = 8.0\text{ V}$. (The battery voltage splits in proportion to the resistances.)

$\dfrac{1}{R} = \dfrac{1}{6.0} + \dfrac{1}{3.0} = \dfrac{1}{6} + \dfrac{2}{6} = \dfrac{3}{6} = \dfrac{1}{2}$, so $R = 2.0\ \Omega$. (The parallel total is always smaller than the smallest resistor, so $9\ \Omega$, from adding them, is wrong.)

Using $P = \dfrac{V^2}{R}$: $R = \dfrac{V^2}{P} = \dfrac{240^2}{60} = \dfrac{57600}{60} = 960\ \Omega$. (Or find the current $I = P/V = 0.25\text{ A}$ first, then $R = V/I = 960\ \Omega$.)

Energy used $= 2.0\text{ kW} \times 3.0\text{ h} = 6.0\text{ kWh}$. Cost $= 6.0 \times \$0.30 = \$1.80$. (The kilowatt-hour is energy: power in kW multiplied by time in hours.)

The parallel pair combines to $\dfrac{6 \times 3}{6 + 3} = 2.0\ \Omega$. In series with the $4.0\ \Omega$ resistor the total is $4.0 + 2.0 = 6.0\ \Omega$. So $I = \dfrac{V}{R} = \dfrac{12}{6.0} = 2.0\text{ A}$. (Reduce the parallel section first, then treat it as a simple series circuit.)

Charge passed: $q = It = 0.50 \times 60 = 30\text{ C}$. Number of electrons: $n = \dfrac{q}{e} = \dfrac{30}{1.6\times10^{-19}} \approx 1.9\times10^{20}$. (Two steps: current times time for the charge, then divide by the electron charge.)

Resistance is $R = \dfrac{\rho L}{A}$. Doubling the length doubles $R$; doubling the diameter multiplies the cross-sectional area by $2^2 = 4$, which divides $R$ by 4. The net effect is $R \times \dfrac{2}{4} = \dfrac{1}{2}R$, so the resistance halves.

In parallel, each globe is connected directly across the full battery voltage, independent of the other. Removing one globe leaves the remaining globe with the same voltage across it, so its current and brightness are unchanged. (In series, removing one globe would break the circuit and the other would go out.)

In a series (voltage divider) circuit the voltage splits in the ratio of the resistances: $V_4 = 9.0 \times \dfrac{4.0}{2.0 + 4.0} = 9.0 \times \dfrac{4}{6} = 6.0\text{ V}$. (The larger resistor gets the larger share of the voltage.)

By the right-hand grip rule, point your right thumb along the current (out of the page, toward you) and your fingers curl in the field direction: anticlockwise. The field forms concentric circles around the wire, getting weaker further away.

$B = \dfrac{\mu_0 I}{2\pi r} = \dfrac{4\pi\times10^{-7} \times 10}{2\pi \times 0.050} = \dfrac{2\times10^{-7} \times 10}{0.050} = \dfrac{2\times10^{-6}}{0.050} = 4.0\times10^{-5}\text{ T}$. (The $\pi$ cancels, and $5.0\text{ cm}$ = $0.050\text{ m}$.)

Inside a long solenoid the field from all the loops adds together to give a strong, nearly uniform field running along the axis, just like the field inside a bar magnet. The field emerges from one end (the north pole) and loops around outside to the other end.

Flux is $\Phi = BA\cos\theta$, where $\theta$ is the angle between the field and the coil's normal (perpendicular). If the coil's plane is parallel to the field, the field lines skim along the surface and none pass through: $\theta = 90^\circ$, $\cos 90^\circ = 0$, so $\Phi = 0$. (Maximum flux occurs when the plane is perpendicular to the field.)

The field of a straight wire is $B = \dfrac{\mu_0 I}{2\pi r} \propto \dfrac{I}{r}$. Doubling both $I$ and $r$ gives $\dfrac{2I}{2r} = \dfrac{I}{r}$, unchanged. The two effects exactly cancel.

The turns per metre is $n = \dfrac{500}{0.25} = 2000\text{ /m}$. Then $B = \mu_0 n I = 4\pi\times10^{-7} \times 2000 \times 2.0 \approx 5.0\times10^{-3}\text{ T}$. (Use turns per unit length, not the total number of turns.)

First the circuit gives the current: $I = \dfrac{V}{R} = \dfrac{6.0}{3.0} = 2.0\text{ A}$. Then the solenoid field: $B = \mu_0 n I = 4\pi\times10^{-7} \times 1000 \times 2.0 \approx 2.5\times10^{-3}\text{ T}$. (This links a circuit calculation to a magnetism one.)