AP Physics 2 Magnetism and Electromagnetic Induction — Worked Answer Explanations

Unit 4 · 14% of the AP exam · 12 questions explained

1. Question 1 · Easy

   A proton moves with velocity $v$ in a magnetic field $B$. The magnetic force on the proton is given by:

   • A
     $F=qv\cdot B$ (dot product)

     Why not A: The magnetic force involves a cross product, not a dot product. The dot product would give a scalar.
   • B
     $F=qv\times B$Correct
   • C
     $F=qB\times v$

     Why not C: The order matters in a cross product: $B\times v=-(v\times B)$, giving the wrong direction.
   • D
     $F=qvB$ (parallel to $v$)

     Why not D: The magnetic force is always perpendicular to $v$, never parallel to it.
   Explanation

   The Lorentz magnetic force on a charge $q$ moving with velocity $v$ in field $B$ is:
   $$F=qv\times B$$
   The magnitude is $F=qvB\sin \theta$ where $\theta$ is the angle between $v$ and $B$. Key properties:

   • Force is always perpendicular to both $v$ and $B$
   • Force does no work on the charge (can change direction but not speed)
   • Direction found by right-hand rule (for positive charges)
   Key takeaway

   Magnetic force: $\vec{F} = q\vec{v} \times \vec{B}$; always perpendicular to velocity, does no work.

2. Question 2 · Easy

   Lenz's law states that the direction of an induced current is such that it:

   • A
     Enhances the change in magnetic flux that caused it.

     Why not A: Lenz's law says the induced current opposes the change — if it enhanced the change, it would violate conservation of energy.
   • B
     Is always clockwise when viewed from above.

     Why not B: The direction depends on the specific change in flux, not a fixed clockwise rule.
   • C
     Opposes the change in magnetic flux that caused it.Correct
   • D
     Is proportional to the magnitude of the magnetic field.

     Why not D: The current magnitude depends on the rate of flux change and circuit resistance, not the field magnitude alone.
   Explanation

   Lenz's law is the physical interpretation of the minus sign in Faraday's law ($\mathcal{E}=-d{\Phi }_B/dt$). The induced current flows in a direction that produces a magnetic field opposing the change in flux:

   • If flux is increasing, the induced current creates a field opposing the increase (trying to keep ${\Phi }_B$ constant).
   • If flux is decreasing, the induced current creates a field that maintains the flux.

   This is a consequence of energy conservation — the induced current acts as a 'brake' against the cause that produced it.

   Key takeaway

   Lenz's law: induced current opposes the change in flux; this is the physical content of the minus sign in Faraday's law.

3. Question 3 · Easy

   A long straight wire carries current $I$ to the right. A proton (positive charge) is located directly above the wire and is moving to the right (same direction as $I$). What is the direction of the magnetic force on the proton?

   • A
     Upward, away from the wire

     Why not A: The field above the wire points out of the page; the force on a rightward-moving positive charge in an out-of-page field is downward, not upward.
   • B
     Downward, toward the wireCorrect
   • C
     To the left, opposing the proton's motion

     Why not C: The magnetic force is always perpendicular to velocity; it cannot have a component along or opposite the velocity direction.
   • D
     Out of the page

     Why not D: Out of the page is the direction of the magnetic field above the wire, not the force on the proton.
   Explanation

   Step 1: Magnetic field of the wire. Using the right-hand rule for a wire carrying current to the right (+x): above the wire, the field points out of the page (+z direction).

   Step 2: Force on the proton. Proton moves in +x direction, field is in +z direction:
   $$F=qv\times B=q(v\hat{x})\times (B\hat{z})=qvB(\hat{x}\times \hat{z})=qvB(-\hat{y})$$

   The force is in the $-y$ direction (downward), toward the wire. Parallel currents attract — this is consistent with the proton moving in the same direction as the current.

   Key takeaway

   Above a wire with rightward current, $\vec{B}$ points out of page; a rightward-moving positive charge is pulled downward (toward the wire).

4. Question 4 · Easy

   A rectangular loop of wire is placed in a uniform magnetic field $B$ pointing into the page. The loop lies in the plane of the page. If the magnetic flux through the loop is ${\Phi }_B=BA$ where $A$ is the loop area, what is the induced EMF when the loop is stationary?

   • A
     $\mathcal{E}=BA$

     Why not A: The EMF is the rate of change of flux, not the flux itself; constant flux gives zero EMF.
   • B
     $\mathcal{E}=B/A$

     Why not B: This has no physical basis in Faraday's law.
   • C
     $\mathcal{E}=0$Correct
   • D
     $\mathcal{E}=B^{2}A$

     Why not D: This has no basis in Faraday's law and incorrect dimensions.
   Explanation

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

   If the loop is stationary and the field $B$ is uniform and constant, then ${\Phi }_B=BA=\text{constant}$, so:
   $$\mathcal{E}=-\frac{d{\Phi }_B}{dt}=0$$

   No EMF is induced. EMF requires a changing magnetic flux — either a changing $B$, changing area $A$, or changing angle between $B$ and the loop's normal vector.

   Key takeaway

   Faraday's law: $\mathcal{E} = -d\Phi_B/dt$; a stationary loop in a constant field has zero EMF (no changing flux).

5. Question 5 · Medium

   A circular loop of radius $r=0.1\mathrm{m}$ has resistance $R=5\Omega$. It is placed in a magnetic field perpendicular to the loop. The field changes at a rate of $\frac{dB}{dt}=200T/s$. What is the induced current in the loop?

   • A
     $I\approx 0.04\mathrm{A}$

     Why not A: This uses $r$ alone instead of the area $\pi r^2$: $200\times 0.1/5=4$… off by the missing $\pi r$ factor.
   • B
     $I\approx 1.26\mathrm{A}$Correct
   • C
     $I\approx 6.28\mathrm{A}$

     Why not C: This is the induced EMF ($\mathcal{E}=\pi r^2\cdot dB/dt\approx 6.28\mathrm{V}$) without dividing by $R=5\Omega$ to obtain current.
   • D
     $I\approx 0.4\mathrm{A}$

     Why not D: This substitutes the diameter ($2r=0.2\mathrm{m}$) instead of the area $\pi r^2$ into Faraday's law.
   Explanation

   Apply Faraday's law to find the induced EMF magnitude:
   $$|\mathcal{E}|=A\frac{dB}{dt}=\pi r^2\frac{dB}{dt}=\pi (0.1{)}^2\times 200=\pi \times 0.01\times 200=2\pi \approx 6.28\mathrm{V}$$

   Induced current via Ohm's law:
   $$I=\frac{|\mathcal{E}|}{R}=\frac{6.28}{5}\approx 1.26\mathrm{A}$$

   By Lenz's law, the current flows in the direction that opposes the increase in flux. The key steps are: (1) compute the area $A=\pi r^2$, (2) multiply by $dB/dt$ to get EMF, (3) divide by $R$ to get current.

   Key takeaway

   Induced EMF: $\mathcal{E} = \pi r^2(dB/dt)$ for a circular loop; induced current $I = \mathcal{E}/R$.

6. Question 6 · Medium

   A proton moves in a circle of radius $r$ when it enters a region with a uniform magnetic field $B$ perpendicular to its velocity. Which of the following expressions gives the speed of the proton? ($m_p$ = proton mass, $e$ = proton charge)

   • A
     $v=\frac{eB}{m_{p}r}$

     Why not A: This inverts the $r$ factor; from $qvB=m{v}^2/r$, solving for $v$ gives $v=qBr/m$, not $qB/(mr)$.
   • B
     $v=\frac{eBr}{m_p}$Correct
   • C
     $v=\frac{m_{p}r}{eB}$

     Why not C: This inverts the entire expression, putting $m_{p}r$ in the numerator instead of the denominator.
   • D
     $v=eBr{m}_p$

     Why not D: This multiplies all quantities instead of using the correct ratio from Newton's second law.
   Explanation

   A charged particle in a uniform magnetic field moves in a circle. The magnetic force provides the centripetal force:
   $$qvB=\frac{m{v}^2}{r}$$
   Solving for $v$ (cancel one $v$):
   $$qB=\frac{mv}{r}\Rightarrow v=\frac{qBr}{m}=\frac{eBr}{m_p}$$

   This is the cyclotron radius equation (rearranged). The radius is $r=mv/(qB)$ — heavier particles and faster particles have larger radii; stronger fields produce smaller radii.

   Key takeaway

   Circular motion in magnetic field: $qvB = mv^2/r$, so $v = qBr/m$ and $r = mv/(qB)$.

7. Question 7 · Medium

   A rectangular conducting loop is pulled out of a uniform magnetic field $B$ pointing into the page at constant velocity $v$. The loop has width $L$ (the side perpendicular to $v$). Which expression gives the magnitude of the induced EMF while the loop is being pulled out?

   • A
     $\mathcal{E}=B{L}^{2}v$

     Why not A: This uses $L^2$ instead of $L$; only one dimension ($L$, the width) enters the motional EMF formula.
   • B
     $\mathcal{E}=BLv$Correct
   • C
     $\mathcal{E}=Bv/L$

     Why not C: This divides by $L$ instead of multiplying; the longer the side in the field, the greater the EMF.
   • D
     $\mathcal{E}=B^{2}Lv$

     Why not D: EMF is linear in $B$, not $B^2$; the extra $B$ factor has no basis in Faraday's law.
   Explanation

   As the loop is pulled out, the area inside the field decreases. The rate of change of flux:
   $$\mathcal{E}=-\frac{d{\Phi }_B}{dt}=-B\frac{dA}{dt}$$

   As the loop exits at speed $v$, the area inside the field decreases at rate $dA/dt=-Lv$ (where $L$ is the width). Therefore:
   $$|\mathcal{E}|=BLv$$

   This is the motional EMF formula for a conductor of length $L$ moving at speed $v$ in field $B$. By Lenz's law, the induced current flows in a direction to oppose the decrease in flux (into page), so the induced current creates a field into the page inside the loop — meaning clockwise current as viewed from above.

   Key takeaway

   Motional EMF when pulling a loop of width $L$ out of a field at speed $v$: $\mathcal{E} = BLv$.

8. Question 8 · Medium

   Two long parallel wires carry equal currents $I$ in opposite directions and are separated by distance $d$. What is the nature of the magnetic force between them?

   • A
     Attractive, because opposite currents create fields that reinforce between the wires.

     Why not A: Fields reinforce between the wires for opposite currents, but the force on each wire is determined by the field at its location due to the other wire — which gives a repulsive force.
   • B
     Repulsive, because the magnetic fields between the wires push outward.Correct
   • C
     Attractive, same as for parallel currents in the same direction.

     Why not C: Parallel currents in the same direction attract; opposite currents repel.
   • D
     There is no magnetic force because the fields cancel between the wires.

     Why not D: Even if fields partially cancel in the region between wires, each wire still experiences a force from the field produced by the other wire.
   Explanation

   Using the right-hand rule:

   • Wire 1 (current upward) creates a field that circles counterclockwise (viewed from above). At wire 2's location (to the right), this field points into the page.
   • Wire 2 carries current downward in the field pointing into the page: $F=IL\times B$. With $I$ downward and $B$ into the page, the force on wire 2 is to the right (away from wire 1).

   Result: opposite currents repel; same-direction currents attract. The rule: "parallel means attract, anti-parallel means repel" for current-carrying wires.

   Key takeaway

   Parallel wires with opposite currents repel; parallel wires with same-direction currents attract — use $\vec{F} = I\vec{L} \times \vec{B}$.

9. Question 9 · Hard

   A square loop of side $s=0.2\mathrm{m}$ and resistance $R=4\Omega$ is in a magnetic field that is increasing at $dB/dt=50T/s$ perpendicular to the loop. What is the power dissipated in the loop?

   • A
     $P=0.25\mathrm{W}$

     Why not A: This computes $I^2={\mathcal{E}}^2/R^2=4/16=0.25$ but forgets to multiply by $R$ to get power; this gives only $I^2$, not $P=I^{2}R$.
   • B
     $P=1.0\mathrm{W}$Correct
   • C
     $P=16\mathrm{W}$

     Why not C: This uses $P={\mathcal{E}}^2\times R=4\times 4$ instead of $P={\mathcal{E}}^2/R$; the formula has $R$ in the denominator.
   • D
     $P=0.5\mathrm{W}$

     Why not D: This gives $P=\mathcal{E}\times I=2\times 0.5/2$, or uses $\mathcal{E}/(2R)$ — a sign/factor error in the power formula.
   Explanation

   Step 1: Area of square loop: $A=s^2=(0.2{)}^2=0.04{\mathrm{m}}^2$

   Step 2: Induced EMF (Faraday's law):
   $$\mathcal{E}=A\frac{dB}{dt}=0.04{\mathrm{m}}^2\times 50T/s=2.0\mathrm{V}$$

   Step 3: Induced current:
   $$I=\frac{\mathcal{E}}{R}=\frac{2.0}{4}=0.5\mathrm{A}$$

   Step 4: Power dissipated:
   $$P=I^{2}R=(0.5{)}^2\times 4=0.25\times 4=1.0\mathrm{W}$$

   Alternatively: $P={\mathcal{E}}^2/R=(2.0{)}^2/4=4/4=1.0\mathrm{W}$ ✓

   Key takeaway

   Power in an induced circuit: $P = \mathcal{E}^2/R = I^2R$; first find $\mathcal{E} = A(dB/dt)$, then $P = \mathcal{E}^2/R$.

10. Question 10 · Hard

    A long solenoid has $n$ turns per unit length and carries current $I$. Which expression gives the magnetic field inside the solenoid?

    • A
      $B={\mu }_{0}I/(2\pi r)$

      Why not A: This is the field of a long straight wire at distance $r$, not the interior field of a solenoid.
    • B
      $B={\mu }_{0}nI$Correct
    • C
      $B={\mu }_{0}NI/r$ where $N$ is total turns and $r$ is radius

      Why not C: This incorrectly introduces the radius $r$; solenoid interior field is uniform and does not depend on radius.
    • D
      $B={\mu }_{0}n{I}^2$

      Why not D: The field is linear in $I$, not proportional to $I^2$.
    Explanation

    Derived from Ampere's law applied to a rectangular Amperian loop straddling the solenoid wall:
    $$B\cdot l={\mu }_0(n\cdot l)I$$
    $$\boxed{B={\mu }_{0}nI}$$
    where $n$ is turns per unit length (turns/m) and $I$ is the current. Key properties:

    • Field inside is uniform (independent of position)
    • Field outside is approximately zero
    • Analogous to the uniform field between capacitor plates but for magnetic fields

    The long straight wire formula $B={\mu }_{0}I/(2\pi r)$ applies at distance $r$ from the wire — a completely different geometry.

    Key takeaway

    Inside a solenoid: $B = \mu_0 n I$ (uniform); outside: $B \approx 0$. Don't confuse with the straight-wire formula.

11. Question 11 · Hard

    A conducting rod of length $L=0.5\mathrm{m}$ slides along frictionless rails at constant velocity $v=4m/s$ in a uniform magnetic field $B=0.3\mathrm{T}$ perpendicular to the plane of the rails. The resistance of the circuit is $R=0.6\Omega$. What external force is needed to maintain constant velocity?

    • A
      $F=0.15\mathrm{N}$Correct
    • B
      $F=0$

      Why not B: Frictionless rails need no force to overcome friction, but the magnetic braking force requires an applied force to maintain constant speed.
    • C
      $F=0.6\mathrm{N}$

      Why not C: This may be the induced current ($I=BLv/R=0.3\times 0.5\times 4/0.6=1\mathrm{A}$) without multiplying by $BL$ to get force.
    • D
      $F=0.30\mathrm{N}$

      Why not D: This may result from computing $BLv$ alone without dividing by $R$ to get current first.
    Explanation

    Step 1: Motional EMF: $\mathcal{E}=BLv=0.3\times 0.5\times 4=0.6\mathrm{V}$

    Step 2: Induced current: $I=\mathcal{E}/R=0.6/0.6=1\mathrm{A}$

    Step 3: Braking force on rod (from $F=IL\times B$): $F_{\text{brake}}=BIL=0.3\times 1\times 0.5=0.15\mathrm{N}$ (opposing motion, by Lenz's law)

    Step 4: To maintain constant velocity, applied force must equal the braking force:
    $$F_{\text{applied}}=F_{\text{brake}}=0.15\mathrm{N}$$

    Alternatively, using power: $P=F_{\text{applied}}\cdot v={\mathcal{E}}^2/R=0.36/0.6=0.6\mathrm{W}$, so $F=P/v=0.6/4=0.15\mathrm{N}$ ✓

    Key takeaway

    Sliding rod: $\mathcal{E} = BLv$, $I = \mathcal{E}/R$, braking force $F = BIL$; applied force matches braking force for constant velocity.

12. Question 12 · Hard

    An electron enters a region with both a uniform electric field $E=1000V/m$ pointing upward and a uniform magnetic field $B=0.02\mathrm{T}$ pointing out of the page. The electron moves horizontally to the right without deflection. What must be the speed of the electron?

    • A
      $v=5\times 10^{4}m/s$Correct
    • B
      $v=2\times 10^{-5}m/s$

      Why not B: This inverts the ratio $E/B$; the selector equation is $v=E/B$, not $B/E$.
    • C
      $v=5\times 10^{4}m/s$ in a specific direction only

      Why not C: The velocity selector condition $v=E/B$ is valid for this specific geometry; the direction is already fixed by the problem (horizontal right).
    • D
      $v=2\times 10^{7}m/s$

      Why not D: This uses $E\times B=1000\times 0.02=20$, which has wrong units; $v=E/B$ not $v=E\times B$.
    Explanation

    For the electron to travel undeflected, the electric force and magnetic force must balance.

    Electric force on electron ($-e$, field $E$ upward): Force is downward (force on negative charge is opposite to $E$): $F_E=eE$ (downward)

    Magnetic force on electron (velocity rightward $+x$, $B$ out of page $+z$):
    $$F_B=qv\times B=(-e)(v\hat{x})\times (B\hat{z})=-evB(\hat{x}\times \hat{z})=-evB(-\hat{y})=evB\hat{y}$$
    Magnetic force is upward on the electron.

    For equilibrium (no deflection):
    $$F_E=F_B\Rightarrow eE=evB\Rightarrow v=\frac{E}{B}$$
    $$v=\frac{1000}{0.02}=5\times 10^{4}m/s$$

    This is a velocity selector — only particles with $v=E/B$ pass straight through, regardless of their charge or mass.

    Key takeaway

    Velocity selector: balance electric and magnetic forces, giving $v = E/B$; works for any charged particle at this speed.