AP Physics C: Electricity and Magnetism Magnetic Fields — Worked Answer Explanations

Unit 4 · 12 questions explained

1. Question 1 · Easy

   A proton ($q=+e$, $m_p$) moves with velocity $v=v\hat{x}$ through a magnetic field $B=B\hat{z}$. The magnetic force on the proton is:

   • A
     $F=-evB\hat{y}$, directed in the $-y$ direction.Correct
   • B
     $F=+evB\hat{y}$, directed in the $+y$ direction.

     Why not B: $\hat{x}\times \hat{z}=-\hat{y}$ (not $+\hat{y}$). Confirm with right-hand rule: point fingers in $+\hat{x}$, curl toward $+\hat{z}$, and the thumb points in $-\hat{y}$.
   • C
     $F=+evB\hat{z}$, directed along the magnetic field.

     Why not C: The magnetic force is always perpendicular to $B$. A force parallel to $B$ is impossible from the cross product $qv\times B$.
   • D
     $F=0$, because the proton moves perpendicular to the field.

     Why not D: $F=0$ only when $v$ is parallel to $B$. When $v\perp B$, the magnitude is maximum: $F=qvB\sin 90°=qvB$.
   Explanation

   The magnetic (Lorentz) force: $F=qv\times B=e(v\hat{x})\times (B\hat{z})=evB(\hat{x}\times \hat{z})$.

   Using the right-hand rule or the determinant: $\hat{x}\times \hat{z}=-(\hat{z}\times \hat{x})=-\hat{y}$.

   $$F=evB(-\hat{y})=-evB\hat{y}$$

   The force is in the $-y$ direction. The proton curves in the $xy$-plane.

   Key takeaway

   $\vec{F} = q\vec{v}\times\vec{B}$; use right-hand rule. $\hat{x}\times\hat{z} = -\hat{y}$.

2. Question 2 · Easy

   A long straight wire carries current $I=5\text{A}$. The magnetic field at a perpendicular distance $r=0.10\text{m}$ from the wire has magnitude (using ${\mu }_0=4\pi \times 10^{-7}\text{T⋅m/A}$):

   • A
     $B=10\mu \text{T}$Correct
   • B
     $B=100\mu \text{T}$

     Why not B: This is 10 times too large. Check: $B=(4\pi \times 10^{-7})(5)/(2\pi \times 0.10)=10^{-5}\text{T}=10\mu \text{T}$, not $100\mu \text{T}$.
   • C
     $B=1\mu \text{T}$

     Why not C: This is 10 times too small, possibly from using $r=1\text{m}$ rather than $r=0.10\text{m}$.
   • D
     $B=50\mu \text{T}$

     Why not D: This may result from using $B={\mu }_{0}I/r$ (omitting the $2\pi$): $(4\pi \times 10^{-7})(5)/(0.10)\approx 63\mu \text{T}$. Ampère's law for a wire gives ${\mu }_{0}I/(2\pi r)$.
   Explanation

   From Ampère's law for an infinite straight wire:
   $$B=\frac{{\mu }_{0}I}{2\pi r}=\frac{(4\pi \times 10^{-7})(5)}{2\pi (0.10)}=\frac{20\pi \times 10^{-7}}{0.2\pi }=\frac{20\times 10^{-7}}{0.2}=10\times 10^{-6}\text{T}=10\mu \text{T}$$
   The field circles the wire (right-hand rule: thumb along current, fingers curl in direction of $B$).

   Key takeaway

   Straight wire: $B = \mu_0 I/(2\pi r)$. Don't forget the $2\pi$ in the denominator.

3. Question 3 · Easy

   Using the Biot-Savart law, the magnetic field at the center of a circular loop of radius $R$ carrying current $I$ is:

   • A
     $B=\frac{{\mu }_{0}I}{2R}$Correct
   • B
     $B=\frac{{\mu }_{0}I}{4\pi R}$

     Why not B: This is ${\mu }_{0}I/(4\pi R)$, which would result from integrating $dB$ over only a quarter circle. The full circle gives ${\mu }_{0}I/(2R)$.
   • C
     $B=\frac{{\mu }_{0}I}{2\pi R}$

     Why not C: This is the field of a straight wire at distance $R$, not a circular loop. The loop integral gives ${\mu }_{0}I/(2R)$ — larger than the straight-wire result.
   • D
     $B=\frac{{\mu }_{0}I{R}^2}{2(R^2+R^2{)}^{3/2}}$

     Why not D: This applies the on-axis loop formula at $z=R$, not at the center ($z=0$). At the center, use $z=0$ in $B_{axis}={\mu }_{0}I{R}^2/[2(R^2+z^2{)}^{3/2}]$ to get ${\mu }_{0}I/(2R)$.
   Explanation

   Biot-Savart: $dB=\frac{{\mu }_0}{4\pi }\frac{Idl\times \hat{r}}{r^2}$.

   At the center of a loop, every element $dl$ is perpendicular to $\hat{r}$ (radial), and all contributions point in the same direction (along the axis). The distance is $r=R$ everywhere:
   $$dB=\frac{{\mu }_0}{4\pi }\frac{Idl}{R^2}$$
   $$B=\frac{{\mu }_{0}I}{4\pi R^2}\oint dl=\frac{{\mu }_{0}I}{4\pi R^2}(2\pi R)=\frac{{\mu }_{0}I}{2R}$$

   Key takeaway

   Circular loop center: $B = \mu_0 I/(2R)$ via Biot-Savart. All $dl$ elements contribute equally in the same direction.

4. Question 4 · Easy

   A solenoid of length $L=0.50\text{m}$, $N=1000$ turns, and radius $r\ll L$ carries current $I=2\text{A}$. Using Ampère's law, the magnetic field inside the solenoid is:

   • A
     $B={\mu }_{0}nI\approx 5.0\text{mT}$Correct
   • B
     $B={\mu }_{0}NI/(2\pi r)$

     Why not B: This is the field formula for a toroid, not a solenoid. A long solenoid with $n$ turns/meter gives $B={\mu }_{0}nI$.
   • C
     $B={\mu }_{0}NI/L^2$

     Why not C: The Ampère path for a solenoid gives $B={\mu }_0(N/L)I={\mu }_{0}nI$, not ${\mu }_{0}NI/L^2$. Dividing by $L^2$ gives incorrect units.
   • D
     $B={\mu }_{0}I/(2\pi r)$

     Why not D: This is the field at distance $r$ from a single straight wire. A solenoid is not a single wire; the field is ${\mu }_{0}nI$ independent of radius inside.
   Explanation

   Apply Ampère's law with a rectangular Amperean loop, one side inside the solenoid (length $\ell$) and the opposite side outside (where $B\approx 0$):
   $$B\ell ={\mu }_0{I}_{\text{enc}}={\mu }_0(n\ell )I$$
   $$B={\mu }_{0}nI$$
   where $n=N/L=1000/0.50=2000$ turns/m.
   $$B=(4\pi \times 10^{-7})(2000)(2)=4\pi \times 10^{-7}\times 4000\approx 5.03\times 10^{-3}\text{T}=5.0\text{mT}$$

   Key takeaway

   Solenoid interior: $B = \mu_0 nI$ (uniform, axial). Derived from Ampère's law with rectangular path.

5. Question 5 · Medium

   A rectangular current loop ($a\times b$) carrying current $I$ is placed in a uniform magnetic field $B$ with its plane parallel to $B$ (the normal to the loop is perpendicular to $B$). The torque on the loop is:

   • A
     $\tau =IAB$Correct
   • B
     $\tau =0$

     Why not B: Torque is zero when the normal to the loop is parallel to $B$ (equilibrium position), not when the plane is parallel to $B$. With the plane parallel to $B$, $\theta =90°$ and $\tau =\mu B\sin 90°=IAB$.
   • C
     $\tau =2IAB$

     Why not C: There is no factor of 2. The torque formula is $\tau =\mu \times B$, where $|\mu |=IA$. The maximum torque is $IAB$ when $\theta =90°$.
   • D
     $\tau =I{A}^{2}B/a$

     Why not D: This incorrectly introduces a length ratio. The torque depends on the area $A=ab$ and current, not on the individual side lengths separately.
   Explanation

   The magnetic dipole moment of the loop: $\mu =IA\hat{n}$, where $A=ab$ and $\hat{n}$ is the normal to the loop.

   Torque: $\tau =\mu \times B$, so $|\tau |=\mu B\sin \theta$.

   When the plane of the loop is parallel to $B$, the normal $\hat{n}$ is perpendicular to $B$, so $\theta =90°$:
   $$\tau =IAB\sin 90°=IAB$$

   This is the maximum torque on the loop. The torque tends to rotate the loop so that $\hat{n}$ aligns with $B$.

   Key takeaway

   $\tau = \mu B\sin\theta = IAB\sin\theta$; maximum when loop plane is parallel to $\vec{B}$ (i.e., $\theta = 90°$).

6. Question 6 · Medium

   Using Ampère's law, the magnetic field at a distance $r$ inside a long cylindrical conductor of radius $R$ carrying a total current $I$ uniformly distributed over its cross section is:

   • A
     $B=\frac{{\mu }_{0}Ir}{2\pi R^2}$ for $r<R$Correct
   • B
     $B=\frac{{\mu }_{0}I}{2\pi r}$ for $r<R$

     Why not B: This is the correct formula for the field outside the conductor ($r>R$). Inside, only the enclosed current (proportional to $r^2$) contributes.
   • C
     $B=0$ for $r<R$

     Why not C: Zero field inside applies to a hollow conducting shell or the interior of an ideal solenoid — not a solid conductor with current flowing through it.
   • D
     $B=\frac{{\mu }_{0}I}{2\pi R^2}$ for $r<R$

     Why not D: This is missing the factor of $r$ in the numerator. Inside, $I_{\text{enc}}=I(r/R{)}^2$, giving $B\propto r$, not a constant.
   Explanation

   For $r<R$, enclosed current (uniform current density $J=I/\pi R^2$):
   $$I_{\text{enc}}=J\pi r^2=\frac{I}{\pi R^2}\cdot \pi r^2=\frac{I{r}^2}{R^2}$$

   Ampère's law with circular Amperean loop of radius $r$:
   $$B(2\pi r)={\mu }_0{I}_{\text{enc}}=\frac{{\mu }_{0}I{r}^2}{R^2}$$
   $$B=\frac{{\mu }_{0}Ir}{2\pi R^2}$$

   For $r>R$: $B={\mu }_{0}I/(2\pi r)$. Field is maximum at $r=R$: $B_{\max }={\mu }_{0}I/(2\pi R)$.

   Key takeaway

   Inside solid wire: $B = \mu_0 Ir/(2\pi R^2)$ — linear in $r$. Outside: $B = \mu_0 I/(2\pi r)$. Match at $r = R$.

7. Question 7 · Medium

   A particle of mass $m$, charge $q$, enters a uniform magnetic field $B$ perpendicular to its velocity $v$. The radius of the circular orbit is:

   • A
     $r=\frac{mv}{qB}$Correct
   • B
     $r=\frac{qB}{mv}$

     Why not B: This inverts the correct expression. Equating magnetic force to centripetal: $qvB=m{v}^2/r$ gives $r=mv/(qB)$, not $qB/(mv)$.
   • C
     $r=\frac{m{v}^2}{qB}$

     Why not C: An extra factor of $v$ appears in the numerator. The centripetal relation $qvB=m{v}^2/r$ gives $r=mv/(qB)$, not $m{v}^2/(qB)$.
   • D
     $r=\frac{m}{qvB}$

     Why not D: This places $v$ in the denominator. From $qvB=m{v}^2/r$: $r=mv/(qB)$; velocity appears in the numerator.
   Explanation

   The magnetic force provides centripetal acceleration:
   $$qvB=\frac{m{v}^2}{r}$$
   Solving for $r$:
   $$r=\frac{mv}{qB}$$
   This is the cyclotron radius (or Larmor radius). The period of circular orbit:
   $$T=\frac{2\pi r}{v}=\frac{2\pi m}{qB}$$
   is independent of speed — the basis of the cyclotron.

   Key takeaway

   Circular motion in $B$: $r = mv/(qB)$. Period $T = 2\pi m/(qB)$ — independent of speed.

8. Question 8 · Medium

   Two long parallel wires separated by distance $d$ carry currents $I_1$ and $I_2$ in the same direction. The force per unit length between them is:

   • A
     $F/L=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}$, attractive.Correct
   • B
     $F/L=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}$, repulsive.

     Why not B: Parallel currents in the same direction attract each other. (Antiparallel currents repel.) The magnitude is correct but the direction is wrong.
   • C
     $F/L=\frac{{\mu }_0{I}_1{I}_2}{4\pi d}$, attractive.

     Why not C: The factor $4\pi$ appears in Coulomb's law and Biot-Savart, but Ampère's result for a wire gives ${\mu }_{0}I/(2\pi r)$ — the denominator is $2\pi d$.
   • D
     $F/L=\frac{{\mu }_0{I}_1{I}_2}{\pi d^2}$, repulsive.

     Why not D: The force falls as $1/d$ (from the $1/d$ field of one wire), not $1/d^2$. And parallel currents attract.
   Explanation

   Field from wire 1 at the location of wire 2:
   $$B_1=\frac{{\mu }_0{I}_1}{2\pi d}$$
   Force per unit length on wire 2 (current $I_2$, length $L$) in field $B_1$:
   $$\frac{F}{L}=I_2{B}_1=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}$$

   Direction (right-hand rule): field from wire 1 curls toward wire 2; force on wire 2 is $I_{2}l\times B_1$, directed toward wire 1 → attractive.

   This is how the SI ampere was historically defined.

   Key takeaway

   $F/L = \mu_0 I_1 I_2/(2\pi d)$; parallel same-direction currents attract, antiparallel repel.

9. Question 9 · Medium

   Using the Biot-Savart law, the magnetic field at a point $P$ on the axis of a circular loop of radius $R$ at axial distance $z$ from the center is:

   • A
     $B=\frac{{\mu }_{0}I{R}^2}{2(R^2+z^2{)}^{3/2}}$Correct
   • B
     $B=\frac{{\mu }_{0}I}{2\sqrt{R^2+z^2}}$

     Why not B: This omits the factor $R^2$ in the numerator and the correct $3/2$ power. The axial component of $dB$ includes a geometric factor $R/(\sqrt{R^2+z^2})$, giving $R^2/(R^2+z^2{)}^{3/2}$ overall.
   • C
     $B=\frac{{\mu }_{0}IR}{2(R^2+z^2)}$

     Why not C: The denominator should be $(R^2+z^2{)}^{3/2}$, not $(R^2+z^2{)}^1$. The extra factor of $(R^2+z^2{)}^{1/2}$ comes from projecting $dB$ onto the axis.
   • D
     $B=\frac{{\mu }_{0}I}{2R}$ (same as at the center, independent of $z$).

     Why not D: The field at the center ($z=0$) is ${\mu }_{0}I/(2R)$. Away from the center it decreases as the denominator $(R^2+z^2{)}^{3/2}$ grows. The field is not constant along the axis.
   Explanation

   By Biot-Savart, each element $dl$ of the loop is at distance $\ell =\sqrt{R^2+z^2}$ from $P$. The element field is perpendicular to $r$. By symmetry, only the axial component survives:
   $$d{B}_{\text{axis}}=\frac{{\mu }_0}{4\pi }\frac{Idl}{R^2+z^2}\cdot \frac{R}{\sqrt{R^2+z^2}}$$
   Integrating around the full loop ($\oint dl=2\pi R$):
   $$B=\frac{{\mu }_{0}I}{4\pi }\frac{2\pi R\cdot R}{(R^2+z^2{)}^{3/2}}=\frac{{\mu }_{0}I{R}^2}{2(R^2+z^2{)}^{3/2}}$$
   At $z=0$: $B={\mu }_{0}I/(2R)$. For $z\gg R$: $B\approx {\mu }_{0}I{R}^2/(2z^3)={\mu }_0\mu /(2\pi z^3)$ (dipole field).

   Key takeaway

   On-axis loop field: $B = \mu_0 IR^2/[2(R^2+z^2)^{3/2}]$. At center ($z=0$): $\mu_0 I/(2R)$.

10. Question 10 · Hard

    A Hall-effect sensor: a flat conductor (width $w$, thickness $t$) carries current $I$ in the $+x$ direction in a field $B=B\hat{z}$. Charge carriers are electrons (charge $-e$, number density $n$). At steady state, the magnitude of the Hall voltage across width $w$ is:

    • A
      $V_H=\frac{IB}{net}$Correct
    • B
      $V_H=\frac{IB}{2net}$

      Why not B: There is no factor of 2. The Hall field exactly balances the magnetic force: $e{E}_H=e{v}_{d}B$, and with $J=ne{v}_d=I/(wt)$, we get $V_H=E_{H}w=v_{d}Bw=IB/(net)$.
    • C
      $V_H=0$, because the magnetic force on electrons and ions cancels.

      Why not C: The Hall effect specifically separates charges. In a conductor with only one carrier type (electrons), there is no cancellation — electrons accumulate on one edge, creating a voltage.
    • D
      $V_H=\frac{IB}{2net}$ (split evenly between edges).

      Why not D: There is no factor of 2 in the Hall voltage. The Hall field exactly balances the magnetic force on charges at steady state: $E_H=v_{d}B$, giving $V_H=IB/(net)$.
    Explanation

    Electrons drift in $-x$ direction (opposite to conventional current). Magnetic force on an electron:
    $$F=(-e)v_d\times B=(-e)(-v_d\hat{x})\times (B\hat{z})=e{v}_{d}B(\hat{x}\times \hat{z})=-e{v}_{d}B\hat{y}$$
    Force is in $-y$, so electrons pile up at the $-y$ edge (bottom). This makes the bottom edge negative and the top edge positive.

    At steady state, the Hall electric field $E_H$ (pointing in $-y$) balances magnetic force:
    $$e{E}_H=e{v}_{d}B\Rightarrow E_H=v_{d}B$$
    Current density $J=ne{v}_d$, so $v_d=I/(newt)$:
    $$V_H=E_{H}w=v_{d}Bw=\frac{IB}{net}$$

    The sign convention: electrons accumulate at bottom → bottom is at lower potential → $V_H=V_{\text{top}}-V_{\text{bottom}}>0$, but the question's sign depends on the reference edge.

    Key takeaway

    Hall voltage: $V_H = IB/(net)$. Electrons drift opposite to $I$; magnetic force pushes them to one edge, setting up $E_H$.

11. Question 11 · Hard

    A toroid has $N$ turns, inner radius $a$, outer radius $b$, and carries current $I$. Using Ampère's law with a circular Amperean loop of radius $r$ ($a<r<b$), the magnetic field inside the toroid is:

    • A
      $B=\frac{{\mu }_{0}NI}{2\pi r}$Correct
    • B
      $B={\mu }_{0}nI$ where $n=N/(2\pi r)$, same as a straight solenoid.

      Why not B: This is effectively the same as choice A (both give ${\mu }_{0}NI/(2\pi r)$), but the key difference from a straight solenoid is that the toroid field depends on $r$ (non-uniform), whereas a solenoid field is uniform. Choice A is the explicit formula.
    • C
      $B=\frac{{\mu }_{0}NI}{2\pi a}$ (uniform inside the toroid).

      Why not C: The toroid field varies with $r$ — it is stronger near the inner radius $a$ and weaker near the outer radius $b$. The field is not uniform (unlike an infinite solenoid).
    • D
      $B=0$ inside the toroid.

      Why not D: Zero field applies to the region outside the toroid ($r>b$ or $r<a$), not inside. Inside, the Amperean loop encloses $NI$ total current.
    Explanation

    Apply Ampère's law with a circular Amperean loop of radius $r$ (concentric with the toroid, $a<r<b$):
    $$\oint B\cdot dl=B(2\pi r)={\mu }_0{I}_{\text{enc}}={\mu }_{0}NI$$
    $$B=\frac{{\mu }_{0}NI}{2\pi r}$$

    Key points:

    • For $r<a$ (inside the toroid hole): $I_{\text{enc}}=0$ → $B=0$.
    • For $r>b$ (outside): each turn's current is threaded twice (in and out) → $I_{\text{enc}}=0$ → $B=0$.
    • The field is azimuthal and non-uniform inside, stronger closer to the inner radius.
    Key takeaway

    Toroid: $B = \mu_0 NI/(2\pi r)$ inside; zero outside. Unlike a solenoid, the field varies with $r$ (non-uniform).

12. Question 12 · Hard

    A proton ($m_p=1.67\times 10^{-27}\text{kg}$, $q=1.60\times 10^{-19}\text{C}$) moving with speed $v=2.0\times 10^6\text{m/s}$ enters a region with both $E$ and $B$ fields. For the proton to travel in a straight line (velocity selector), with $v=v\hat{x}$, $B=B\hat{z}=0.050\text{T}\hat{z}$, the required electric field $E$ and its magnitude are:

    • A
      $E=1.0\times 10^5\text{V/m}(-\hat{y})$

      Why not A: The magnetic force on the proton is $F_B=qvB(-\hat{y})$, so the balancing electric force must be in $+\hat{y}$, requiring $E$ in $+\hat{y}$. A field in $-\hat{y}$ would add to the magnetic deflection rather than cancel it.
    • B
      $E=1.0\times 10^5\text{V/m}(+\hat{y})$Correct
    • C
      $E=1.0\times 10^5\text{V/m}(+\hat{x})$

      Why not C: An electric field in $+\hat{x}$ would accelerate the proton along its direction of motion, not balance the transverse magnetic force. The balance must be in the $y$-direction.
    • D
      $E=4.0\times 10^{-14}\text{V/m}(+\hat{y})$

      Why not D: This uses $E=m_{p}v/(qL)$ with some length $L$, which has no basis. The balance condition is simply $qE=qvB$, giving $E=vB=1.0\times 10^5\text{V/m}$.
    Explanation

    For the proton to travel in a straight line, the net force must be zero: $F_E+F_B=0$.

    Magnetic force: $F_B=qv\times B=q(v\hat{x})\times (B\hat{z})=qvB(\hat{x}\times \hat{z})=qvB(-\hat{y})$.

    For balance: $qE=-F_B=qvB(+\hat{y})$, so $E=vB\hat{y}$.

    $$|E|=vB=(2.0\times 10^6)(0.050)=1.0\times 10^5\text{V/m}$$

    This is a velocity selector: only particles with $v=E/B$ pass through undeflected, regardless of mass or charge sign.

    Key takeaway

    Velocity selector: $E = vB$ for straight-line travel; $\vec{E}$ must oppose magnetic force. Selects speed, not mass.