Electromagnetic Induction

Faraday's law is $\varepsilon = -N\dfrac{\Delta\Phi}{\Delta t}$. Only a changing flux induces an EMF; a constant flux, no matter how large, gives $\varepsilon = 0$.

When the plane is perpendicular to the field, the normal is parallel to it, so $\theta = 0$: $\Phi = BA\cos 0 = (0.50)(0.020) = 0.010\text{ Wb}$.

$\Phi = BA\cos\theta$, where $\theta$ is the angle between the coil's normal and the field. Flux is maximum when the normal lies along $B$, i.e. the coil's plane is perpendicular to the field so the lines pass straight through the face.

$|\varepsilon| = N\dfrac{|\Delta\Phi|}{\Delta t} = 50 \times \dfrac{0.020}{0.10} = 50 \times 0.20 = 10\text{ V}$.

The flux is increasing, so by Lenz's law the induced current opposes it: the near face becomes a north pole and repels the magnet. You must do work pushing it in, and that work becomes electrical energy (conservation of energy).

$\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$, so $V_s = 240 \times \dfrac{100}{1200} = 20\text{ V}$. Fewer secondary turns means a step-down transformer.

Ideal transformer conserves power: $V_p I_p = V_s I_s$, so $I_p = \dfrac{V_s I_s}{V_p} = \dfrac{(20)(6.0)}{240} = \dfrac{120}{240} = 0.50\text{ A}$. Stepping voltage down steps current up.

An ideal transformer conserves power, $V_p I_p = V_s I_s$. Raising the voltage by some factor lowers the current by the same factor. You cannot get more power out than you put in.

A transformer relies on a changing flux linking the secondary. Steady DC gives a constant flux in the core, so $\dfrac{\Delta\Phi}{\Delta t} = 0$ and $\varepsilon = -N\dfrac{\Delta\Phi}{\Delta t} = 0$. Transformers work only on AC.

Slip rings keep each coil end on its own ring, giving sinusoidal AC. A split-ring commutator swaps the connections every half turn, flipping the negative half-cycles up to give pulsed (rectified) DC. Same generator, different contacts.

First $\omega = 2\pi f = 2\pi(25) = 157\text{ rad/s}$. Then $\varepsilon_{\max} = NBA\omega = (200)(0.40)(0.015)(157) = 1.2 \times 157 \approx 190\text{ V}$.

Plane parallel to the field means the normal is perpendicular ($\theta = 90^\circ$), so $\Phi_i = 0$. Plane perpendicular means $\Phi_f = BA = (0.30)(0.020) = 0.0060\text{ Wb}$. Then $|\varepsilon| = N\dfrac{\Delta\Phi}{\Delta t} = 100 \times \dfrac{0.0060}{0.050} = 12\text{ V}$.

$\eta = \dfrac{4700}{5000} = 94\%$. The core heats mainly through eddy currents; laminating it (thin insulated sheets) breaks up those current loops. Thicker copper cuts winding $I^2R$ loss, not core loss.

EMF depends on the rate of change of flux, not the flux itself. With $\Phi = BA\cos(\omega t)$, the flux changes fastest as it passes through zero, when the coil's plane is parallel to the field. Maximum flux is where the EMF is momentarily zero, a quarter-cycle out of phase.

The rod sweeps out area at rate $Lv$, so $\varepsilon = BLv = (0.50)(0.40)(3.0) = 0.60\text{ V}$. (Equivalently $\varepsilon = \dfrac{\Delta\Phi}{\Delta t} = B\dfrac{\Delta A}{\Delta t}$.)

Current: $I = \dfrac{\varepsilon}{R} = \dfrac{0.60}{0.15} = 4.0\text{ A}$. This current in the field feels a retarding force $F = BIL = (0.50)(4.0)(0.40) = 0.80\text{ N}$ (Lenz's law opposes the motion), so at constant speed the applied force equals $0.80\text{ N}$.

The moving magnet changes the flux through the pipe walls, inducing circular eddy currents (Faraday). By Lenz's law these currents oppose the change, dragging on the magnet. Copper is not magnetic; the braking is purely induction, converting gravitational energy into heat in the pipe.

For a fixed power $P$, the line current is $I = \dfrac{P}{V}$, so the line loss is $P_{\text{line}} = I^2R = \left(\dfrac{P}{V}\right)^2 R \propto \dfrac{1}{V^2}$. Ten times the voltage means $10^2 = 100$ times less loss. This is why transformers step voltage up for transmission.

$\varepsilon_{\max} = NBA\omega = NBA(2\pi f) \propto f$, and the output completes one cycle per rotation so its frequency equals the spin frequency. Doubling the speed doubles both.

First the area: $A = \pi r^2 = \pi(0.050)^2 = 7.9\times10^{-3}\text{ m}^2$. With $N$ turns square-on ($\theta = 0$): $\varepsilon = NA\dfrac{\Delta B}{\Delta t} = (40)(7.9\times10^{-3})(2.0) \approx 0.63\text{ V}$. Forgetting the $40$ turns gives the trap $0.016\text{ V}$.

Flipping reverses the normal, so the flux goes from $+BA$ to $-BA$: the change is $|\Delta\Phi| = 2BA = 2(0.25)(0.010) = 0.0050\text{ Wb}$ (not $BA$). Then $|\varepsilon| = N\dfrac{|\Delta\Phi|}{\Delta t} = 80 \times \dfrac{0.0050}{0.040} = 10\text{ V}$.

The rotor is driven by currents induced in its bars. If it matched the field's speed there would be no relative motion, so the flux through the bars would not change, no current would be induced, and there would be no torque. It must "slip" behind to keep inducing current.

First the field's synchronous speed: $n_s = \dfrac{120f}{P} = \dfrac{120(50)}{4} = 1500\text{ rpm}$. Then $\text{slip} = \dfrac{n_s - n_r}{n_s} = \dfrac{1500 - 1440}{1500} = 0.04 = 4\%$. A few percent slip is typical.

The output is $96\%$ of the input, so $P_{\text{in}} = \dfrac{4800}{0.96} = 5000\text{ W}$. The waste heat (eddy currents, hysteresis, winding $I^2R$) is $5000 - 4800 = 200\text{ W}$.

$\varepsilon_{\max} = NBA\omega$. The frequency fixes $\omega = 2\pi f$, so spinning faster would change the frequency. To double $\varepsilon_{\max}$ at fixed $f$, double one of the other factors: $N$, $B$ or $A$.