Electromagnetic Induction — Hard Practice Quiz
A Physics cheat sheet for Electromagnetic Induction — every key formula with its symbols defined — plus a hard-level practice quiz to test recall.
Formulas & key concepts
Magnetic Flux: Flux \(\Phi\) through area \(A\) in magnetic field \(B\). \(\phi\) is angle between normal and field.
Faraday's Law: Induced EMF \(\mathcal{E}\) in a coil of \(N\) turns is proportional to rate of change of magnetic flux.
Motional EMF: EMF induced in a conductor of length \(L\) moving with speed \(v\) perpendicular to magnetic field \(B\).
Electric Generator: EMF induced in a rotating coil. Peak EMF \(\mathcal{E}_0 = NAB\omega\).
Mutual Inductance: EMF induced in coil 2 due to changing current in coil 1. \(M\) is mutual inductance.
Self Inductance: EMF induced in a coil due to changing current in itself. \(L\) is self-inductance.
Energy Stored in Inductor: Magnetic potential energy stored in an inductor carrying current \(I\).
Transformers: Relationship between primary (p) and secondary (s) voltages, turns, and currents.
Practice quiz
A square loop of side $L$ is placed in a uniform magnetic field $B$ perpendicular to the plane of the loop. If the loop is stretched to a square of side $2L$ in time $\Delta t$, while the magnetic field remains constant, what is the magnitude of the average induced EMF?
- $\frac{3BL^2}{\Delta t}$
- $\frac{BL^2}{\Delta t}$
- $\frac{4BL^2}{\Delta t}$
- $\frac{2BL^2}{\Delta t}$
Answer: $\frac{3BL^2}{\Delta t}$
An electric generator with $N$ turns, area $A$, and magnetic field $B$ produces a peak EMF of $\mathcal{E}_0$ when rotating at angular frequency $\omega$. If the number of turns is doubled, the area is halved, and the angular frequency is also doubled, what is the new peak EMF in terms of $\mathcal{E}_0$?
- $2\mathcal{E}_0$
- $\mathcal{E}_0$
- $4\mathcal{E}_0$
- $\frac{1}{2}\mathcal{E}_0$
Answer: $2\mathcal{E}_0$
A conducting rod of length $L$ moves with constant velocity $v$ perpendicular to a uniform magnetic field $B$. If the rod is part of a closed circuit with constant resistance $R$, and the magnetic field strength is suddenly doubled while the rod's velocity remains constant, how does the induced current in the circuit change?
- The induced current doubles.
- The induced current halves.
- The induced current quadruples.
- The induced current remains unchanged.
Answer: The induced current doubles.
An inductor stores $U$ joules of energy when a current $I$ flows through it. If the current is increased to $2I$ and the self-inductance is simultaneously halved, what is the new energy stored in the inductor?
- $2U$
- $U$
- $4U$
- $\frac{1}{2}U$
Answer: $2U$
A step-down transformer has a primary coil with $N_p$ turns and a secondary coil with $N_s$ turns, where $N_p = 10 N_s$. If the primary voltage is $V_p$ and the primary current is $I_p$, what is the power dissipated in the secondary circuit, assuming an ideal transformer?
- $V_p I_p$
- $10 V_p I_p$
- $\frac{1}{10} V_p I_p$
- $100 V_p I_p$
Answer: $V_p I_p$
A circular coil of $N$ turns and radius $r$ is placed in a uniform magnetic field $B$ such that the field lines are perpendicular to the coil's plane. If the magnetic field changes uniformly from $B_1$ to $B_2$ in time $\Delta t$, derive an expression for the magnitude of the induced EMF in terms of $N$, $r$, $B_1$, $B_2$, and $\Delta t$.
- $N \frac{|B_2 - B_1| \pi r^2}{\Delta t}$
- $N \frac{(B_2 + B_1) \pi r^2}{\Delta t}$
- $N \frac{|B_2 - B_1| \pi r}{\Delta t}$
- $N \frac{B_1 \pi r^2}{\Delta t}$
Answer: $N \frac{|B_2 - B_1| \pi r^2}{\Delta t}$
A conducting rod of length $L$ moves at a constant speed $v$ on two parallel conducting rails in a uniform magnetic field $B$ perpendicular to the plane of the rails. The rails are connected by a resistor $R$. What is the rate at which mechanical energy is converted into electrical energy in the circuit?
- $\frac{(vBL)^2}{R}$
- $vBLR$
- $\frac{vBL}{R}$
- $\frac{vB^2 L^2}{R^2}$
Answer: $\frac{(vBL)^2}{R}$
A coil has a self-inductance $L$. If the current through the coil is changing at a rate of $\frac{\Delta I}{\Delta t}$, an EMF $\mathcal{E}$ is induced. If the number of turns in the coil is doubled while keeping the coil's geometry (area, length) and the rate of change of current constant, how does the self-inductance $L$ and the induced EMF $\mathcal{E}$ change? (Assume self-inductance $L$ is proportional to the square of the number of turns, $N^2$).
- $L$ becomes $4L$, $\mathcal{E}$ becomes $4\mathcal{E}$.
- $L$ becomes $2L$, $\mathcal{E}$ becomes $2\mathcal{E}$.
- $L$ becomes $4L$, $\mathcal{E}$ becomes $2\mathcal{E}$.
- $L$ becomes $2L$, $\mathcal{E}$ becomes $4\mathcal{E}$.
Answer: $L$ becomes $4L$, $\mathcal{E}$ becomes $4\mathcal{E}$.
Two coils, A and B, are placed near each other. When the current in coil A changes at a rate of $\frac{\Delta I_A}{\Delta t}$, an EMF $\mathcal{E}_B$ is induced in coil B due to mutual inductance $M$. If the current in coil B changes at the same rate, $\frac{\Delta I_B}{\Delta t} = \frac{\Delta I_A}{\Delta t}$, what is the relationship between the magnitude of the EMF induced in coil A ($\mathcal{E}_A$) and $\mathcal{E}_B$?
- $|\mathcal{E}_A| = |\mathcal{E}_B|$
- $|\mathcal{E}_A| = M^2 |\mathcal{E}_B|$
- $|\mathcal{E}_A| = \frac{1}{M} |\mathcal{E}_B|$
- $|\mathcal{E}_A|$ depends on the self-inductance of coil A, not $M$.
Answer: $|\mathcal{E}_A| = |\mathcal{E}_B|$
An AC generator produces a peak EMF $\mathcal{E}_0$. If the angular frequency $\omega$ is doubled, and the magnetic field $B$ is halved, how does the maximum power output of the generator change, assuming it is connected to a constant resistance $R$?
- The maximum power output remains unchanged.
- The maximum power output doubles.
- The maximum power output halves.
- The maximum power output quadruples.
Answer: The maximum power output remains unchanged.
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