The Timpson Lyceum · University Physics II (2026) – Electricity & Magnetism

Chapter 29 — Electromagnetic Induction

Problems with worked solutions · 14 points · 3 extra credit

Chapter 29 — Electromagnetic Induction

Core: 14 points · Extra credit: 3 points. Show your reasoning; partial credit is given.

1. 1. (1 pt •) A straight conducting rod of length $L=0.40\,\text{m}$ slides at constant speed $v=6.0\,\text{m/s}$ along two frictionless rails in a uniform magnetic field $B=0.25\,\text{T}$ perpendicular to the plane of the rails. Find the motional EMF induced across the ends of the rod.

   Solution

   1.

   Motional EMF: $\mathcal{E}=BLv=(0.25)(0.40)(6.0)\approx 0.60\,\text{V}$.

2. 2. (2 pts ••) A flat circular coil has $N=200$ turns and radius $r=3.0\,\text{cm}$. It sits with its plane perpendicular to a magnetic field that increases uniformly from $0$ to $0.50\,\text{T}$ in $0.020\,\text{s}$.

   (a) Find the magnetic flux through a single turn at the instant the field reaches $0.50\,\text{T}$.

   (b) Find the magnitude of the EMF induced in the coil while the field is changing.

   Solution

   2.

   (a) Flux through one turn at the final field:

   $\Phi_B = BA = B\,\pi r^2 = (0.50)\,\pi(0.030)^2 \approx 1.4\times10^{-3}\,\text{Wb}.$

   (b) The field ramps uniformly, so $dB/dt = 0.50/0.020 = 25\,\text{T/s}$. With the plane perpendicular to $\vec B$,

   $|\mathcal{E}| = N A\,\frac{dB}{dt} = (200)(\pi(0.030)^2)(25) \approx 14\,\text{V}.$

3. 3. (2 pts ••) For each situation, state the direction of the induced current in the loop and justify it with Lenz's law.

   (a) A circular loop lies flat on a table. A bar magnet is lowered toward it from above, north pole pointing down. Viewed from above, is the induced current clockwise or counterclockwise?

   (b) The same magnet is now pulled straight back up, away from the loop. Clockwise or counterclockwise (viewed from above)?

   (c) A rectangular loop moves at constant velocity entirely within a large uniform magnetic field (the loop never crosses a field boundary). Is there any induced current? Explain.

   Solution

   3.

   (a) Lowering the north pole downward increases the downward flux through the loop. By Lenz's law the induced current opposes the increase, so it must create an upward field inside the loop. By the right-hand rule that requires a counterclockwise current as viewed from above.

   (b) Pulling the magnet away decreases the downward flux. The induced current now opposes the decrease by reinforcing the downward field, requiring a clockwise current as viewed from above (opposite to part (a)).

   (c) No induced current. The field is uniform and the loop stays entirely within it, so the flux $\Phi_B = BA$ through the loop never changes. With $d\Phi_B/dt = 0$, Faraday's law gives $\mathcal{E}=0$. (The motional EMFs on the two sides parallel to the motion are equal and cancel around the loop.)

4. 4. (2 pts ••) A square loop of side $a=0.15\,\text{m}$ sits in a uniform field $B=0.40\,\text{T}$. Initially the loop's normal is parallel to the field. The loop is then rotated so that its normal makes an angle of $60^\circ$ with the field, taking $0.10\,\text{s}$ to do so.

   (a) Find the magnetic flux through the loop before and after the rotation.

   (b) Find the average EMF induced during the rotation.

   Solution

   4.

   (a) $\Phi_B = BA\cos\theta$ with $A=a^2=(0.15)^2=0.0225\,\text{m}^2$.

   Before ($\theta=0$): $\Phi_i = (0.40)(0.0225)(1) = 9.0\times10^{-3}\,\text{Wb}$.

   After ($\theta=60^\circ$): $\Phi_f = (0.40)(0.0225)(0.5) = 4.5\times10^{-3}\,\text{Wb}$.

   (b) $|\mathcal{E}_{\text{avg}}| = \dfrac{|\Delta\Phi_B|}{\Delta t} = \dfrac{9.0\times10^{-3}-4.5\times10^{-3}}{0.10} \approx 0.045\,\text{V}$.

5. 5. (3 pts •••) A conducting rod of length $L=0.50\,\text{m}$ and mass $m=0.020\,\text{kg}$ rides on horizontal frictionless rails of total circuit resistance $R=2.0\,\Omega$ in a vertical uniform field $B=0.60\,\text{T}$. A constant horizontal force $F=0.18\,\text{N}$ is applied along the rails, and the rod eventually reaches a terminal (constant) speed.

   (a) Explain physically why the rod approaches a terminal speed rather than accelerating forever.

   (b) Find the terminal speed $v_t$.

   (c) At terminal speed, find the induced current in the rod and verify that the electrical power dissipated equals the mechanical power delivered by the applied force.

   Solution

   5.

   (a) As the rod speeds up, the motional EMF $BLv$ and hence the induced current $I=BLv/R$ grow. The magnetic force on that current, $F_{\text{ind}}=BIL=B^2L^2v/R$, opposes the motion (Lenz's law) and grows with $v$. The rod stops accelerating when this retarding force balances the applied force; that balance defines the terminal speed.

   (b) Set $F = \dfrac{B^2L^2 v_t}{R}$:

   $v_t = \frac{FR}{B^2L^2} = \frac{(0.18)(2.0)}{(0.60)^2(0.50)^2} = \frac{0.36}{0.09} = 4.0\,\text{m/s}.$

   (c) Induced current: $I = \dfrac{BLv_t}{R} = \dfrac{(0.60)(0.50)(4.0)}{2.0} = 0.60\,\text{A}.$

   Electrical power dissipated: $P_{\text{elec}} = I^2R = (0.60)^2(2.0) = 0.72\,\text{W}.$

   Mechanical power delivered: $P_{\text{mech}} = Fv_t = (0.18)(4.0) = 0.72\,\text{W}.$

   They match, as required by energy conservation at terminal speed.

6. 6. (2 pts ••) A long solenoid of radius $R=6.0\,\text{cm}$ carries a current that is increasing so that the uniform field inside grows at $dB/dt=0.20\,\text{T/s}$. The changing field induces a circulating electric field.

   (a) Find the magnitude of the induced electric field at a radius $r=4.0\,\text{cm}$ from the axis (inside the solenoid).

   (b) Find the magnitude of the induced electric field at $r=10.0\,\text{cm}$ from the axis (outside the solenoid).

   Solution

   6.

   The induced electric field circulates in loops concentric with the axis. Applying $\oint \vec E\cdot d\vec\ell = -\,d\Phi_B/dt$ to a circle of radius $r$:

   (a) Inside ($r<R$), the enclosed flux is $\Phi_B=\pi r^2 B$, so $E(2\pi r)=\pi r^2\,dB/dt$:

   $E = \frac{r}{2}\frac{dB}{dt} = \frac{0.040}{2}(0.20) \approx 4.0\times10^{-3}\,\text{V/m}.$

   (b) Outside ($r>R$), only the flux through the solenoid's cross-section ($\pi R^2$) changes, so $E(2\pi r)=\pi R^2\,dB/dt$:

   $E = \frac{R^2}{2r}\frac{dB}{dt} = \frac{(0.060)^2}{2(0.10)}(0.20) \approx 3.6\times10^{-3}\,\text{V/m}.$

7. 7. (2 pts ••) A parallel-plate capacitor with circular plates of radius $R=5.0\,\text{cm}$ is being charged. Between the plates the electric field is increasing at $dE/dt=2.0\times10^{12}\,\text{V/(m}\cdot\text{s)}$.

   (a) Find the displacement current between the plates.

   (b) Find the magnitude of the magnetic field between the plates at a radius equal to the plate radius, $r=R$.

   Solution

   7.

   (a) Displacement current $I_d = \varepsilon_0\dfrac{d\Phi_E}{dt} = \varepsilon_0 A\dfrac{dE}{dt}$ with $A=\pi R^2=\pi(0.050)^2=7.85\times10^{-3}\,\text{m}^2$:

   $I_d = (8.85\times10^{-12})(7.85\times10^{-3})(2.0\times10^{12}) \approx 0.14\,\text{A}.$

   (b) The displacement current sources a magnetic field just like a real current. Using Ampère's law with a circle of radius $r=R$ enclosing all of $I_d$:

   $B = \frac{\mu_0 I_d}{2\pi R} = \frac{(4\pi\times10^{-7})(0.139)}{2\pi(0.050)} \approx 5.6\times10^{-7}\,\text{T}.$

8. Extra Credit (3 pts •••)

   A flat rectangular loop of $N$ turns and area $A$ is placed in a uniform magnetic field of magnitude $B$. The loop is spun about an axis lying in its plane and perpendicular to $\vec{B}$, rotating at constant angular speed $\omega$. At $t=0$ the loop's normal is aligned with $\vec{B}$. This is the principle of the AC generator.

   (a) Starting from the definition of magnetic flux, write $\Phi_B(t)$ for one turn as a function of time.

   (b) Using Faraday's law, derive the induced EMF $\mathcal{E}(t)$ for the $N$-turn loop, and identify the peak EMF $\mathcal{E}_0$.

   (c) Explain, using your result, at what orientation of the loop the EMF is maximum and at what orientation it is zero — and why this makes physical sense in terms of the rate of change of flux (not the flux itself).

   Solution

   Extra Credit

   (a) The flux through one turn is $\Phi_B = \vec B\cdot\vec A = BA\cos\theta$, where $\theta$ is the angle between the loop's normal and $\vec B$. With constant angular speed and the normal aligned at $t=0$, $\theta=\omega t$, so

   $\Phi_B(t) = BA\cos(\omega t).$

   (b) Faraday's law for $N$ turns:

   $\mathcal{E}(t) = -N\frac{d\Phi_B}{dt} = -N\,BA\frac{d}{dt}\big[\cos(\omega t)\big] = N B A\,\omega\,\sin(\omega t).$

   This is sinusoidal (alternating) with peak value

   $\mathcal{E}_0 = N B A\,\omega.$

   (c) The EMF is largest in magnitude when $\sin(\omega t)=\pm1$, i.e. $\omega t = 90^\circ, 270^\circ$: the loop's plane is parallel to $\vec B$ (normal perpendicular to $\vec B$). At that orientation the instantaneous flux is momentarily zero, but the flux is changing fastest, and Faraday's law depends on $d\Phi_B/dt$, not $\Phi_B$. Conversely, the EMF is zero when the loop's plane is perpendicular to $\vec B$ (normal aligned with $\vec B$): the flux is at its maximum magnitude there, but its rate of change is momentarily zero as it turns around. Peak EMF and peak flux are thus a quarter cycle ($90^\circ$) out of phase.