AP Physics C- Electricity and Magnetism- 13.2 Electromagnetic Induction - Exam Style questions - FRQs- New Syllabus
Question
• At time $t_{1}$ the bar reaches position $x_{1}$ and smoothly transitions to the horizontal section of the rails and enters a uniform magnetic field of magnitude $B_{1}=0.40\text{ T}$ that is directed in the $+y$-direction.
• At time $t_{2}$ the bar reaches position $x_{2}$ and enters a region with no magnetic field.
• At time $t_{3}$, the bar reaches position $x_{3}$ and enters a uniform magnetic field of magnitude $B_{2}=0.60\text{ T}$ that is directed in the $+z$-direction.
• At time $t_{4}$ the bar reaches position $x_{4}$ and enters a region with no magnetic field.
(b) At time $t_{B}$, the speed of the bar is $v=2.5\text{ m/s}$.
ii. Calculate the magnitude of the net force $F_{net}$ exerted on the bar at time $t_{B}$.
(d) The original scenario is repeated but with a new bar that has the same mass but with a nonnegligible resistance $R_{bar}=0.20\text{ }\Omega$. The new bar is released from rest and smoothly transitions to the horizontal section of the rails and enters the first uniform magnetic field.
ii. In the original scenario, the magnitude of the acceleration of the bar immediately after the bar enters the first uniform magnetic field is $a_{original}$. In the new scenario, the magnitude of the acceleration of the bar immediately after the bar enters the first uniform magnetic field is $a_{new}$. Is $a_{new}$ greater than, less than, or equal to $a_{original}$? Justify your answer.
Most-appropriate topic codes (AP Physics C: Electricity and Magnetism):
•Topic 13.2 — Electromagnetic Induction
• Topic 13.3 — Induced Currents and Magnetic Forces
▶️ Answer/Explanation
(a)
Correct direction: An arrow pointing straight in the $-x$ direction (to the left).
As the bar moves in the $+x$ direction through the uniform magnetic field $\vec{B}_1$ pointing in the $+y$ direction, an electromotive force is induced. By Lenz’s law, the magnetic force opposes the relative motion, resulting in a net magnetic field drag directed opposite to the velocity.
(b)(i)
Calculate the induced motional EMF in the bar:
$\mathcal{E} = B_1 v d$
$\mathcal{E} = (0.40\text{ T})(2.5\text{ m/s})(0.30\text{ m}) = 0.30\text{ V}$
Now, apply Ohm’s law to find the magnitude of the electric current:
$I = \frac{\mathcal{E}}{R}$
$I = \frac{0.30\text{ V}}{0.20\text{ }\Omega} = 1.5\text{ A}$
$\boxed{I = 1.5\text{ A}}$
(b)(ii)
Calculate the net magnetic force acting on the current-carrying bar:
$F_{net} = I d B_1$
$F_{net} = (1.5\text{ A})(0.30\text{ m})(0.40\text{ T}) = 0.18\text{ N}$
$\boxed{F_{net} = 0.18\text{ N}}$
(c)
The velocity-time graph should show the following continuous segments:
• From $t = 0$ to $t_1$: A linear increase from zero, since gravity accelerates the bar down the sloped track at constant acceleration.
• From $t_1$ to $t_2$: An exponential decay curve that approaches a constant asymptotic value, representing the magnetic braking force slowing down the bar.
• From $t_2$ to $t_3$: A flat horizontal line with a constant positive velocity, because there is no magnetic field or friction to alter the speed.
• From $t_3$ to $t_4$: A flat horizontal line with the same constant velocity as the previous interval. Because the magnetic field $\vec{B}_2$ is aligned in the $+z$ direction (parallel to the bar length), the cross product $\vec{v} \times \vec{B}$ yields no charge separation along the bar, meaning no EMF or current is induced here.
(d)(i)
The original loop only contains the fixed resistor, but now the bar itself adds extra internal resistance in series:
$R_{total} = R + R_{bar}$
$R_{total} = 0.20\text{ }\Omega + 0.20\text{ }\Omega = 0.40\text{ }\Omega$
$\boxed{R_{total} = 0.40\text{ }\Omega}$
(d)(ii)
Correct selection: Less than.
Immediately upon entering the field region, both bars travel at identical initial velocities since they descended from the same height $H$. However, because the total loop resistance is doubled in the new scenario, the induced current decreases by half ($I = \frac{\mathcal{E}}{R_{total}}$). Consequently, the counter-acting magnetic retarding force ($F = IdB$) diminishes, producing a smaller deceleration magnitude ($a = \frac{F}{m}$).
(e)
To maximize the initial induced current, we must increase the initial motional EMF $\mathcal{E} = B_1 v d$. This can be accomplished by implementing any one of the following changes:
• Increase $H$: Elevating the starting point causes the bar to attain a higher velocity $v = \sqrt{2gH}$ right before entering the field, which boosts the induced EMF.
• Increase $B_1$: A stronger magnetic field directly increases the rate of magnetic flux change as the bar slices across it.
• Increase $d$: A wider track separation increases the active length of the conductor cutting the field lines, raising the motional EMF proportionally.