AP Physics C: Electricity and Magnetism Electromagnetism — Worked Answer Explanations

Unit 5 · 12 questions explained

1. Question 1 · Easy

   A rectangular loop of area $A=0.04{\text{m}}^2$ is in a uniform magnetic field $B$ that increases at rate $dB/dt=50\text{T/s}$. The magnitude of the induced EMF in the loop is:

   • A
     $\mathcal{E}=2.0\text{V}$Correct
   • B
     $\mathcal{E}=0.04\text{V}$

     Why not B: This uses only $A$ and ignores $dB/dt$: $A=0.04$ but $\mathcal{E}=A\cdot dB/dt=0.04\times 50=2.0\text{V}$.
   • C
     $\mathcal{E}=50\text{V}$

     Why not C: This uses only $dB/dt=50\text{T/s}$ and ignores the loop area. Faraday's law requires both: $\mathcal{E}=A\cdot dB/dt$.
   • D
     $\mathcal{E}=1250\text{V}$

     Why not D: This multiplies $A\times (dB/dt{)}^2$ or some incorrect combination. Faraday's law is $\mathcal{E}=|d{\Phi }_B/dt|=AdB/dt$ (for uniform field and fixed area).
   Explanation

   Faraday's law: $\mathcal{E}=-\frac{d{\Phi }_B}{dt}$.

   For a flat loop with area $A$ perpendicular to a uniform field $B$:
   $${\Phi }_B=BA\Rightarrow |\mathcal{E}|=A\frac{dB}{dt}=(0.04)(50)=2.0\text{V}$$

   Lenz's law: the induced current creates a field opposing the increase in $B$.

   Key takeaway

   Faraday's law: $|\mathcal{E}| = |d\Phi_B/dt| = A\,|dB/dt|$ for fixed area perpendicular to uniform field.

2. Question 2 · Easy

   A conducting rod of length $L=0.50\text{m}$ moves with velocity $v=4\text{m/s}$ perpendicular to a uniform magnetic field $B=0.30\text{T}$. The motional EMF induced in the rod is:

   • A
     $\mathcal{E}=BLv=0.60\text{V}$Correct
   • B
     $\mathcal{E}=BL/v=0.0375\text{V}$

     Why not B: The motional EMF is $BLv$ (multiply), not $BL/v$ (divide). Velocity in the denominator has no physical basis.
   • C
     $\mathcal{E}=B/Lv=0.15\text{V}$

     Why not C: This incorrectly divides by both $L$ and $v$. The force on charges is $qvB$, and the EMF across the rod of length $L$ is $\mathcal{E}=vBL$.
   • D
     $\mathcal{E}=B{L}^{2}v=0.30\text{V}$

     Why not D: An extra factor of $L$ appears. The motional EMF is the line integral of $v\times B$ along the rod: $\mathcal{E}=vBL$, with $L$ to the first power.
   Explanation

   A charge $q$ in the moving rod experiences force $F=qv\times B$. The work done per unit charge in moving from one end to the other:
   $$\mathcal{E}={\int }_0^L(v\times B)\cdot dl=vBL$$
   (when $v$, $B$, and $l$ are mutually perpendicular)
   $$\mathcal{E}=(4)(0.30)(0.50)=0.60\text{V}$$

   Key takeaway

   Motional EMF: $\mathcal{E} = BLv$ for rod perpendicular to $\vec{B}$ and $\vec{v}$. It's the work per unit charge by the magnetic force.

3. Question 3 · Easy

   Lenz's law states that the direction of the induced current in a loop is such that:

   • A
     The induced current's magnetic field opposes the change in flux that caused it.Correct
   • B
     The induced current's magnetic field reinforces the change in flux, amplifying the effect.

     Why not B: Reinforcement would violate conservation of energy — a self-amplifying current would require no external energy input. Lenz's law ensures energy conservation by opposing the change.
   • C
     The induced current always flows clockwise when viewed from above.

     Why not C: The direction depends on whether flux is increasing or decreasing and on the orientation of the field. Lenz's law specifies opposition to flux change, not a fixed clockwise or counterclockwise direction.
   • D
     The induced EMF is proportional to the magnetic flux, not its rate of change.

     Why not D: This confuses Faraday's law (EMF $\propto d\Phi /dt$) with a false $\mathcal{E}\propto \Phi$ statement. Lenz's law addresses the sign (direction) of the induced EMF, not its magnitude.
   Explanation

   Lenz's law is the statement of energy conservation embedded in Faraday's law (the negative sign):
   $$\mathcal{E}=-\frac{d{\Phi }_B}{dt}$$

   • If ${\Phi }_B$ is increasing: induced current creates a field opposing the increase (i.e., in the direction to oppose the growing flux).
   • If ${\Phi }_B$ is decreasing: induced current creates a field in the same direction as the original flux (opposing the decrease).

   This is the electromagnetic equivalent of Newton's third law — the system resists changes imposed on it.

   Key takeaway

   Lenz's law: induced current opposes the **change** in flux. Expressed by the minus sign in Faraday's $\mathcal{E} = -d\Phi/dt$.

4. Question 4 · Easy

   A solenoid of $N=500$ turns, length $\ell =0.25\text{m}$, and cross-sectional area $A=2.0\times 10^{-4}{\text{m}}^2$ carries current $I$. The self-inductance of the solenoid is:

   • A
     $L={\mu }_0{n}^{2}A\ell \approx 0.20\text{mH}$Correct
   • B
     $L={\mu }_{0}NA/\ell \approx 0.50\mu \text{H}$

     Why not B: Missing a factor of $N$ in the numerator. Self-inductance is $L={\mu }_0{N}^{2}A/\ell ={\mu }_0{n}^{2}A\ell$, not ${\mu }_{0}NA/\ell$.
   • C
     $L={\mu }_0{N}^2/(A\ell )\approx 1.26\times 10^9\text{H}$

     Why not C: Dividing by $A$ instead of multiplying gives enormous and unphysical inductance. The correct formula has $A$ in the numerator: $L={\mu }_0{N}^{2}A/\ell$.
   • D
     $L={\mu }_0{N}^{2}A\approx 6.28\times 10^{-5}\text{H}$

     Why not D: Missing the division by length $\ell$. Self-inductance depends on the turns-per-unit-length squared: $L={\mu }_0(N/\ell {)}^{2}A\ell ={\mu }_0{N}^{2}A/\ell$.
   Explanation

   Self-inductance from flux linkage $\Lambda =N{\Phi }_B=N(BA)=N({\mu }_{0}nI)A={\mu }_{0}nNIA$:
   $$L=\frac{\Lambda }{I}={\mu }_{0}nNA={\mu }_0\frac{N}{\ell }\cdot N\cdot A=\frac{{\mu }_0{N}^{2}A}{\ell }$$

   With $n=N/\ell =500/0.25=2000\text{turns/m}$:
   $$L={\mu }_0{n}^{2}A\ell =(4\pi \times 10^{-7})(2000{)}^2(2\times 10^{-4})(0.25)$$
   $$=(4\pi \times 10^{-7})(4\times 10^6)(5\times 10^{-5})=4\pi \times 10^{-7}\times 200\approx 2.51\times 10^{-4}\text{H}\approx 0.25\text{mH}$$

   Key takeaway

   Solenoid inductance: $L = \mu_0 N^2 A/\ell = \mu_0 n^2 A\ell$. Increases with $N^2$, area, and decreases with length.

5. Question 5 · Medium

   An RL circuit has $R=50\Omega$, $L=0.10\text{H}$, and battery $\mathcal{E}=20\text{V}$. The switch is closed at $t=0$. The current as a function of time is $I(t)=I_{\max }(1-e^{-t/{\tau }_L})$ with ${\tau }_L=L/R$. The initial rate of change of current $dI/dt{|}_{t=0}$ is:

   • A
     $dI/dt{|}_{t=0}=\mathcal{E}/L=200\text{A/s}$Correct
   • B
     $dI/dt{|}_{t=0}=\mathcal{E}/R=0.40\text{A/s}$

     Why not B: $\mathcal{E}/R$ is the final (maximum) current, not the initial rate of change. At $t=0$ the inductor sets $dI/dt=\mathcal{E}/L$.
   • C
     $dI/dt{|}_{t=0}=0$

     Why not C: Zero initial rate applies to a capacitor (which opposes instantaneous voltage change), not an inductor. An inductor opposes instantaneous current change, but the rate $dI/dt$ starts at its maximum.
   • D
     $dI/dt{|}_{t=0}=\mathcal{E}/(RL)=4\text{A/s}$

     Why not D: Dividing by both $R$ and $L$ introduces wrong units ($\text{A}\cdot {\text{s}}^{-1}$ would require $\mathcal{E}/L$ in units of A/s, not $\mathcal{E}/(RL)$). The KVL at $t=0$ gives $\mathcal{E}=LdI/dt$.
   Explanation

   KVL at $t=0$ (current $I=0$, so $IR=0$):
   E−LdtdI​−IR=0⇒E=LdtdI​​t=0​
   dtdI​​t=0​=LE​=0.1020​=200A/s

   This is the maximum rate of current increase. As $I$ grows, the back-EMF $LdI/dt$ grows until at $t\to \infty$, $dI/dt\to 0$ and $I\to \mathcal{E}/R=0.40\text{A}$.

   Key takeaway

   RL circuit at $t=0$: $dI/dt = \mathcal{E}/L$ (max rate). At $t\to\infty$: $I = \mathcal{E}/R$ (max current).

6. Question 6 · Medium

   An RL circuit ($R=20\Omega$, $L=0.40\text{H}$) is powered by $\mathcal{E}=10\text{V}$. After a long time the switch is opened. The current through the inductor at time $t$ after opening is:

   • A
     $I(t)=I_0{e}^{-Rt/L}=0.50e^{-50t}\text{A}$Correct
   • B
     $I(t)=I_0(1-e^{-Rt/L})$

     Why not B: This is the charging (growth) solution, not the decay. After the switch opens, there is no source, and $I$ decays exponentially from $I_0$.
   • C
     $I(t)=I_0{e}^{-t/RC}$ (RC decay formula)

     Why not C: This is the RC discharge formula. For an RL circuit the time constant is ${\tau }_L=L/R$, not $RC$.
   • D
     $I(t)=0$ immediately after the switch opens.

     Why not D: An inductor cannot change its current instantaneously — that would require infinite voltage. After the switch opens, current decays over time constant $L/R$.
   Explanation

   After a long time (switch closed), the current reaches steady state:
   $$I_0=\mathcal{E}/R=10/20=0.50\text{A}$$

   After the switch opens, KVL gives:
   $$L\frac{dI}{dt}+RI=0\Rightarrow I(t)=I_0{e}^{-Rt/L}$$

   Time constant: ${\tau }_L=L/R=0.40/20=0.020\text{s}$.

   $$I(t)=0.50e^{-20t/0.40}=0.50e^{-50t}\text{A}$$

   The energy stored in the inductor ($\frac{1}{2}L{I}_0^2=0.050\text{J}$) is dissipated in $R$ during the decay.

   Key takeaway

   RL decay: $I(t) = I_0 e^{-Rt/L}$. Current cannot change instantaneously; decays with $\tau = L/R$.

7. Question 7 · Medium

   A circular loop of resistance $R=5\Omega$ and radius $r=0.10\text{m}$ is pulled at constant velocity $v=2\text{m/s}$ out of a region of uniform magnetic field $B=0.4\text{T}$ (perpendicular to the loop). While the loop is partially out, the induced current is:

   • A
     $I_{\text{ind}}=BLv/R$, where $L=2r=0.20\text{m}$ is the chord in the field.

     Why not A: The chord $L=2r$ applies only when the leading edge of the loop has traveled exactly to the center (half-way out). In general the chord varies; this choice applies only at one instant.
   • B
     $I_{\text{ind}}=BLv/R$ where $L$ is the length of the loop edge cutting through the field boundary.Correct
   • C
     $I_{\text{ind}}=B\pi r^{2}v/R$

     Why not C: $\pi r^2$ is the loop area, not the relevant length. The EMF comes from the rate of change of flux: $\mathcal{E}=BLv$ where $L$ is the length of wire crossing the field boundary, not the area.
   • D
     $I_{\text{ind}}=0$, because the flux is decreasing symmetrically.

     Why not D: Flux decreases as the loop exits, producing a non-zero induced EMF and current. Zero current would mean no EMF, which contradicts Faraday's law with changing ${\Phi }_B$.
   Explanation

   As the loop exits the field region at velocity $v$, only the segment of the loop at the field boundary (length $L$ = width of loop at that boundary) contributes motional EMF:
   $$\mathcal{E}=BLv$$
   $$I_{\text{ind}}=\frac{\mathcal{E}}{R}=\frac{BLv}{R}$$

   The direction (Lenz's law): as flux decreases, induced current flows to maintain the flux — by right-hand rule, the current flows counterclockwise when viewed from the field side.

   Note: for a square loop of width $w$ exiting a uniform field, $L=w$ (constant), so EMF is constant during exit. For a circular loop, $L$ varies with position.

   Key takeaway

   Motional EMF from exiting a field: $\mathcal{E} = BLv$ where $L$ is the length of wire at the field boundary. Induced current $= \mathcal{E}/R$.

8. Question 8 · Medium

   The energy stored in an inductor of inductance $L=50\text{mH}$ carrying current $I=4\text{A}$ is:

   • A
     $U_L=\frac{1}{2}L{I}^2=0.40\text{J}$Correct
   • B
     $U_L=L{I}^2=0.80\text{J}$

     Why not B: Missing the factor of $\frac{1}{2}$. Energy stored in an inductor is $\frac{1}{2}L{I}^2$, analogous to kinetic energy $\frac{1}{2}m{v}^2$.
   • C
     $U_L=\frac{L}{2I^2}=0.0016\text{J}$

     Why not C: This inverts $I^2$, placing it in the denominator. Energy increases with current; dividing by $I^2$ would give an inverse relationship.
   • D
     $U_L=\frac{1}{2}\frac{I^2}{L}=160\text{J}$

     Why not D: This divides by $L$ rather than multiplying. Energy is stored in the inductor field and scales with $L$: larger $L$ at the same $I$ stores more energy.
   Explanation

   The work done to build up current in an inductor against the back-EMF:
   $$U_L={\int }_0^I\mathcal{E}{I}^{\prime }dt={\int }_0^{I}L{I}^{\prime }d{I}^{\prime }=\frac{1}{2}L{I}^2$$

   $$U_L=\frac{1}{2}(50\times 10^{-3})(4{)}^2=\frac{1}{2}(0.050)(16)=0.40\text{J}$$

   This energy is stored in the magnetic field inside the inductor (analogous to $\frac{1}{2}C{V}^2$ for a capacitor's electric field).

   Key takeaway

   $U_L = \frac{1}{2}LI^2$. Inductor stores energy in its magnetic field; analogous to $\frac{1}{2}CV^2$ for capacitors.

9. Question 9 · Medium

   Two coils have mutual inductance $M=0.020\text{H}$. Coil 1 carries a current that changes at rate $d{I}_1/dt=30\text{A/s}$. The magnitude of the EMF induced in coil 2 is:

   • A
     $|{\mathcal{E}}_2|=M|d{I}_1/dt|=0.60\text{V}$Correct
   • B
     $|{\mathcal{E}}_2|=M|I_1|=$ unknown (need $I_1$, not $d{I}_1/dt$).

     Why not B: Mutual induction depends on the rate of change of current ($d{I}_1/dt$), not the instantaneous value of $I_1$. The induced EMF is ${\mathcal{E}}_2=-Md{I}_1/dt$.
   • C
     $|{\mathcal{E}}_2|=M^2|d{I}_1/dt|=0.012\text{V}$

     Why not C: Squaring $M$ ($0.020^2=0.00040$) is incorrect. The EMF is linear in $M$: ${\mathcal{E}}_2=Md{I}_1/dt$.
   • D
     $|{\mathcal{E}}_2|=(d{I}_1/dt)/M=1500\text{V}$

     Why not D: This inverts the formula. The induced EMF is $Md{I}_1/dt$ (multiply), not $(d{I}_1/dt)/M$ (divide). Dividing by $M$ gives a very large and physically incorrect result.
   Explanation

   Mutual induction: a changing current in coil 1 creates a changing flux through coil 2, inducing an EMF:
   $${\mathcal{E}}_2=-M\frac{d{I}_1}{dt}$$
   ∣E2​∣=M​dtdI1​​​=(0.020)(30)=0.60V

   This is Faraday's law applied to coupled coils. $M$ depends on geometry and is symmetric: $M_{12}=M_{21}$ — the same $M$ governs the reverse coupling.

   Key takeaway

   $\mathcal{E}_2 = -M\,dI_1/dt$. Mutual inductance $M$ is symmetric; depends on geometry, not current.

10. Question 10 · Hard

    Maxwell's addition to Ampère's law introduces the displacement current $I_d$. For a parallel-plate capacitor being charged, the displacement current between the plates (where there is no real conduction current) is defined as:

    • A
      $I_d={ϵ}_0\frac{d{\Phi }_E}{dt}$Correct
    • B
      $I_d=\frac{1}{{\mu }_0}\frac{d{\Phi }_B}{dt}$

      Why not B: This mixes Faraday's law (changing ${\Phi }_B$ induces $\mathcal{E}$) with the displacement current. Maxwell's term involves $d{\Phi }_E/dt$ (changing electric flux), not changing magnetic flux.
    • C
      $I_d={\mu }_0{ϵ}_0\frac{d{\Phi }_E}{dt}$

      Why not C: The extra factor of ${\mu }_0$ is wrong. Displacement current is $I_d={ϵ}_{0}d{\Phi }_E/dt$, and ${\mu }_0$ appears only when displacement current is substituted into the full Ampère law.
    • D
      $I_d={ϵ}_0\frac{dE}{dt}$ (scalar $E$, not flux).

      Why not D: The displacement current uses the rate of change of electric flux ${\Phi }_E=\int E\cdot dA$, not the rate of change of field magnitude. Dimensionally, ${ϵ}_{0}dE/dt$ has units of A/m${}^2$ (current density), not A.
    Explanation

    Maxwell completed Ampère's law by noting that a changing electric flux also produces a magnetic field:
    $$\oint B\cdot dl={\mu }_0(I+I_d)$$
    where the displacement current is:
    $$I_d={ϵ}_0\frac{d{\Phi }_E}{dt}$$

    For a capacitor being charged with current $I$: the real current flows in the wires, but between the plates $I_d={ϵ}_{0}AdE/dt={ϵ}_0\frac{d}{dt}(Q/{ϵ}_0)=dQ/dt=I$. So the magnetic field is continuous across any surface, whether through the wire or the gap.

    Key takeaway

    Displacement current: $I_d = \epsilon_0 d\Phi_E/dt$. Ensures $\vec{B}$ is continuous around a capacitor and predicts EM waves.

11. Question 11 · Hard

    An LC circuit consists of an inductor $L=0.10\text{H}$ and capacitor $C=100\mu \text{F}$. The capacitor is initially charged to $V_0=10\text{V}$ and the switch is closed at $t=0$. The angular frequency of oscillation $\omega$ and the maximum current $I_{\max }$ are:

    • A
      $\omega =1/\sqrt{LC}\approx 316\text{rad/s}$; $I_{\max }=V_0\sqrt{C/L}\approx 0.316\text{A}$Correct
    • B
      $\omega =1/\sqrt{LC}\approx 316\text{rad/s}$; $I_{\max }=V_0/R=$ undefined (no $R$)

      Why not B: Applying an $R$-based formula to an ideal LC circuit (no resistance) is incorrect. Maximum current occurs when all energy transfers to the inductor: $\frac{1}{2}L{I}_{\max }^2=\frac{1}{2}C{V}_0^2$, giving $I_{\max }=V_0\sqrt{C/L}$.
    • C
      $\omega =\sqrt{LC}\approx 0.00316\text{rad/s}$; $I_{\max }=V_0/\sqrt{LC}\approx 3162\text{A}$

      Why not C: $\omega =\sqrt{LC}$ has wrong units ($\sqrt{\text{H}\cdot \text{F}}=\text{s}$, not rad/s). The correct formula is $\omega =1/\sqrt{LC}$.
    • D
      $\omega =1/(LC)\approx 10^5\text{rad/s}$; $I_{\max }=C{V}_0\omega =$ incorrect

      Why not D: $1/(LC)$ has units of ${\text{F}}^{-1}{\text{H}}^{-1}={\text{s}}^{-2}$, not rad/s. The correct angular frequency is $\omega =1/\sqrt{LC}$.
    Explanation

    The LC circuit equations:
    $$L\frac{dI}{dt}+\frac{Q}{C}=0\Rightarrow \frac{d^{2}Q}{d{t}^2}+\frac{Q}{LC}=0$$
    Solution: $Q(t)=Q_0\cos (\omega t)$ with:
    $$\omega =\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{(0.10)(10^{-4})}}=\frac{1}{\sqrt{10^{-5}}}=\frac{1}{3.16\times 10^{-3}}\approx 316\text{rad/s}$$

    Maximum current (energy conservation):
    $$\frac{1}{2}C{V}_0^2=\frac{1}{2}L{I}_{\max }^2\Rightarrow I_{\max }=V_0\sqrt{\frac{C}{L}}=10\sqrt{\frac{10^{-4}}{0.10}}=10\sqrt{10^{-3}}=10\times 0.0316\approx 0.316\text{A}$$

    Alternately: $I_{\max }=Q_0\omega =C{V}_0/\sqrt{LC}=V_0\sqrt{C/L}$.

    Key takeaway

    LC oscillator: $\omega = 1/\sqrt{LC}$; $I_{\max} = V_0\sqrt{C/L}$ from energy conservation.

12. Question 12 · Hard

    A transformer has primary coil $N_1=200$ turns and secondary coil $N_2=50$ turns. The primary is connected to $V_1=120\text{V}$ (rms) AC. A load $R_L=10\Omega$ is connected to the secondary. Assuming an ideal transformer (no losses), the primary current $I_1$ (rms) is:

    • A
      $I_1=0.75\text{A}$Correct
    • B
      $I_1=12\text{A}$

      Why not B: This would be $V_1/R_L=120/10$, treating the load as directly across the primary. The transformer steps down voltage by $N_2/N_1=1/4$, so $V_2=30\text{V}$ and $I_2=3\text{A}$ — not $I_1$.
    • C
      $I_1=3.0\text{A}$

      Why not C: This is the secondary current $I_2=V_2/R_L=30/10=3\text{A}$, not the primary current. Power conservation gives $I_1=I_2{N}_2/N_1=3\times (50/200)=0.75\text{A}$.
    • D
      $I_1=0.60\text{A}$

      Why not D: This may come from $V_1/R_{\text{primary}}=120/200=0.60$ (using turns as ohms), which has no physical basis. Current is found from power conservation, not from treating $N$ as resistance.
    Explanation

    Step-down transformer ($N_2<N_1$):
    $$V_2=V_1\frac{N_2}{N_1}=120\times \frac{50}{200}=30\text{V}$$

    Secondary current:
    $$I_2=\frac{V_2}{R_L}=\frac{30}{10}=3.0\text{A}$$

    Primary current (from power conservation $V_1{I}_1=V_2{I}_2$):
    $$I_1=\frac{V_2{I}_2}{V_1}=\frac{30\times 3}{120}=\frac{90}{120}=0.75\text{A}$$

    Equivalently: $I_1/I_2=N_2/N_1=1/4$: $I_1=3\times (1/4)=0.75\text{A}$.

    Key takeaway

    Ideal transformer: $V_2/V_1 = N_2/N_1$; $I_1/I_2 = N_2/N_1$. Power $V_1I_1 = V_2I_2$ is conserved.