The Motor Effect

The motor-effect force is $F = BIL\sin\theta$. It is zero when the current is parallel to the field ($\theta = 0$) and greatest when the current is perpendicular. A force needs a component of current across the field.

Use the right-hand rule ($\vec{F} = I\vec{L}\times\vec{B}$): fingers point along the current (right), curl into the field (into the page); the thumb points upward. So the wire is pushed upward.

$F = BIL = (0.50)(3.0)(0.20) = 0.30\text{ N}$, since the current is perpendicular to the field ($\sin 90^\circ = 1$).

$F = BIL\sin\theta = (0.50)(3.0)(0.20)\sin 30^\circ = 0.30 \times 0.5 = 0.15\text{ N}$. Only the component of current perpendicular to the field contributes.

Each wire sits in the magnetic field of the other. When the currents run the same way, the forces pull the wires together: parallel currents attract. (Opposite currents repel.)

$\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d} = \dfrac{(4\pi\times10^{-7})(10)(10)}{2\pi(0.050)} = \dfrac{2\times10^{-7}\times100}{0.050} = 4.0\times10^{-4}\text{ N/m}$.

$\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d} \propto \dfrac{1}{d}$. Doubling the separation $d$ halves the force per unit length.

The forces on opposite sides of the loop are equal and opposite, so they cancel: there is no net force in a uniform field. But those forces act along different lines, forming a couple, so there is a net torque that turns the loop. This torque is what drives a motor.

With the plane parallel to the field the torque is maximum: $\tau = nBIA = (50)(0.20)(2.0)(0.010) = 0.20\text{ N·m}$.

Torque is $\tau = nBIA\cos\theta$, where $\theta$ is the angle between the coil's plane and the field. It is greatest when the plane is parallel to the field, and drops to zero when the plane is perpendicular (face-on), where the side forces have no turning effect.

The upward magnetic force must equal the weight: $BIL = mg$, so $I = \dfrac{mg}{BL} = \dfrac{(0.0040)(9.8)}{(0.10)(0.20)} = \dfrac{0.0392}{0.020} \approx 2.0\text{ A}$.

Rearrange $\tau = nBIA$: $I = \dfrac{\tau}{nBA} = \dfrac{0.30}{(100)(0.25)(0.0060)} = \dfrac{0.30}{0.15} = 2.0\text{ A}$.

Without switching, the coil would reach the vertical position and be pushed back. The commutator reverses the current direction in the coil every half turn, so the force on each side always drives the coil the same way, giving continuous rotation.

With a fixed current, the torque reverses direction after the coil passes the vertical (zero-torque) position, so the coil would swing to vertical and settle there (or oscillate about it) rather than turning continuously. The commutator is what prevents this.

Curved poles make the field point radially, so it is always parallel to the coil's plane as it turns. The coil sides stay perpendicular to the field, keeping the torque $\tau = nBIA$ at its maximum for the whole rotation, giving a smoother, more powerful motor.

First the motor-effect force: $F = BIL = (0.40)(10)(0.50) = 2.0\text{ N}$. Then Newton's second law: $a = \dfrac{F}{m} = \dfrac{2.0}{0.20} = 10\text{ m/s}^2$. (This is the principle of a rail gun.)

The repulsive force per length must support the wire's weight per length: $\dfrac{\mu_0 I^2}{2\pi d} = \left(\dfrac{m}{L}\right)g$. So $I = \sqrt{\dfrac{2\pi d\,(m/L)\,g}{\mu_0}} = \sqrt{\dfrac{2\pi(0.010)(1.0\times10^{-3})(9.8)}{4\pi\times10^{-7}}} \approx 22\text{ A}$.

$\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d} = \dfrac{4\pi\times10^{-7}(1)(1)}{2\pi(1.0)} = 2\times10^{-7}\text{ N/m}$. The ampere was once defined as the current giving exactly this force between such wires.

$F = BIL\sin\theta = (0.30)(4.0)(0.50)\sin 60^\circ = 0.60 \times 0.866 \approx 0.52\text{ N}$.

A spinning coil generates a back EMF that opposes the supply. As the motor speeds up, the back EMF grows, reducing the net voltage $(V - \varepsilon)$ and therefore the current $I = \dfrac{V - \varepsilon}{R}$.

Slowing the motor lowers its back EMF, so the net voltage $(V - \varepsilon)$ and the current $I = \frac{V-\varepsilon}{R}$ both rise. The extra current supplies the extra torque needed for the load, which is also why an overloaded (stalled) motor can overheat.

At switch-on the coil is stationary, so the back EMF is zero and the current is limited only by the resistance: $I = \dfrac{V}{R} = \dfrac{240}{8.0} = 30\text{ A}$. This is the start-up surge.

$V = IR + \varepsilon$, so $\varepsilon = V - IR = 240 - (5.0)(8.0) = 240 - 40 = 200\text{ V}$. The large back EMF at full speed is what limits the running current.

Mechanical power is the back EMF times the current: $P = \varepsilon I = 200 \times 5.0 = 1000\text{ W}$. Efficiency is output over input: $\dfrac{\varepsilon I}{VI} = \dfrac{\varepsilon}{V} = \dfrac{200}{240} \approx 83\%$. The other $I^2R = 200\text{ W}$ is lost as heat.

Before the motor turns there is no back EMF, so the current would be $I = \frac{V}{R}$ with only the small armature resistance, a very large surge. A starter resistor adds temporary resistance to limit this current, then is removed once the growing back EMF holds the current down on its own.