Home / AP Physics C- E&M- 12.4 Ampère’s Law – Exam Style questions – FRQs

AP Physics C- E&M- 12.4 Ampère’s Law - Exam Style questions - FRQs- New Syllabus

AP Physics C - Electricity and Magnetism-Questions - All Topics

Question

A lightbulb of resistance $R = 10.0\ \Omega$ is connected to a rectangular loop of wire of negligible resistance near a very long current-carrying wire. The rectangular loop has a length $L = 4.0\text{ cm}$ and a width $W = 2.0\text{ cm}$ and is positioned so one of the longer sides of the loop is a distance $d = 1.0\text{ cm}$ above and parallel to the long wire, as shown. The current in the long wire is initially flowing to the right and is given by $I(t) = C – Dt$, where $C = 10.0\text{ A}$ and $D = 2.0\text{ A/s}$. At time $t = 5.0\text{ s}$, the current in the long wire is instantaneously zero as the current changes direction.
 
(a) What is the direction, if any, of the magnetic field produced by the induced current in the rectangular loop as the current in the long wire changes direction?
_____ Into the page
_____ Out of the page
_____ No direction, because the field is zero
Justify your answer.
(b) Calculate the magnetic flux through the loop due to only the long wire at time $t = 3.0\text{ s}$.
(c) Calculate the current through the lightbulb at time $t = 3.0\text{ s}$.
(d) A group of students attempts to experimentally verify whether the current through the lightbulb is consistent with the current calculation from part (c). The current in the rectangular loop is measured to be greater than the current calculated in part (c). Which of the following could explain this discrepancy? Select one answer.
_____ The students did not account for Earth’s magnetic field.
_____ The rectangular loop is tilted and is not in the same plane as the wire.
_____ The resistance of the lightbulb is greater than the recorded value.
_____ The long side of the rectangular loop is shorter than the recorded value.
_____ The current in the long wire changes at a faster rate than expected.
Briefly justify your answer.
(e) Later, the same rectangular loop with lightbulb is rotated such that a short side of the loop is $1.0\text{ cm}$ above and parallel to the long current-carrying wire, as shown. The current in the wire is again initially flowing from left to right and given by $I(t) = C – Dt$ where $C = 10.0\text{ A}$ and $D = 2.0\text{ A/s}$. The current through the lightbulb in the loop’s new orientation at time $t = 3.0\text{ s}$ is $I_2$. Which of the following correctly relates the current $I_2$ to $I_1$, the current through the lightbulb in part (c)?
_____ $I_2 < I_1$
_____ $I_2 = I_1$
_____ $I_2 > I_1$
Justify your answer.

Most-appropriate topic codes (AP Physics C: Electricity and Magnetism):

• Topic 12.4 — Ampère’s Law (Parts b, c)
• Topic 13.1 — Magnetic Flux (Parts b, e)
• Topic 13.2 — Electromagnetic Induction (Parts a, c, d, e)
▶️ Answer/Explanation

(a)
Correct selection: Out of the page

Before $t = 5.0\text{ s}$, the current flows to the right, which creates a magnetic field pointing out of the page inside the loop according to the right-hand rule. Because the current is decreasing over time ($D = 2.0\text{ A/s}$), this outward magnetic flux is dropping. Lenz’s law states that the induced current must create its own magnetic field to counteract this decrease, meaning the induced field must also point out of the page to support the fading flux.

(b)
First, find the current in the long wire at $t = 3.0\text{ s}$:

$I(3) = 10.0 – (2.0)(3.0) = 4.0\text{ A}$

The magnetic field at a distance $r$ from a long wire is given by:

$B = \frac{\mu_0 I}{2\pi r}$

Set up the flux integral over the width of the loop using an area element $dA = L \, dr$:

$\Phi_B = \int_d^{d+W} B \, dA = \int_d^{d+W} \frac{\mu_0 I}{2\pi r} L \, dr$

$\Phi_B = \frac{\mu_0 I L}{2\pi} \ln\left(\frac{d+W}{d}\right)$

Substitute the given numerical values into the formula ($d = 0.01\text{ m}$, $W = 0.02\text{ m}$, $L = 0.04\text{ m}$):

\(
\Phi_B = \frac{(4\pi \times 10^{-7})(4.0)(0.04)}{2\pi} \ln\left(\frac{0.01 + 0.02}{0.01}\right)
\)

\(
\Phi_B = (3.2 \times 10^{-8}) \ln(3) \approx 3.52 \times 10^{-8}\text{ T}\cdot\text{m}^2
\)

\(
\boxed{\Phi_B = 3.52 \times 10^{-8}\text{ T}\cdot\text{m}^2}
\)

(c)
Use Faraday’s law to find the magnitude of the induced emf by taking the time derivative of the flux expression:

$\varepsilon = \left|\frac{d\Phi_B}{dt}\right| = \frac{\mu_0 L}{2\pi} \ln\left(\frac{d+W}{d}\right) \left|\frac{dI}{dt}\right|$

Given that \(\left|\frac{dI}{dt}\right| = D = 2.0\text{ A/s}\):

\(\varepsilon = \frac{(4\pi \times 10^{-7})(0.04)}{2\pi} \ln(3) \times 2.0\)

\(\varepsilon = (1.6 \times 10^{-8})\ln(3) \approx 1.76 \times 10^{-8}\text{ V}\)

Now, use Ohm’s law to calculate the current passing through the lightbulb:

\(I_1 = \frac{\varepsilon}{R} = \frac{1.76 \times 10^{-8}}{10.0}\)

\(I_1 = 1.76 \times 10^{-9}\text{ A}\)

\(\boxed{I_1 = 1.76 \times 10^{-9}\text{ A}}\)

(d)
Correct selection: The current in the long wire changes at a faster rate than expected.

If the current changes quicker than predicted, the value of $\left|\frac{dI}{dt}\right|$ is larger than expected. According to Faraday’s law, a higher rate of change in current causes a faster rate of change in the magnetic flux, which directly leads to a larger induced emf and a greater measured current flowing through the loop.

(e)
Correct selection: $I_2 < I_1$

In this new orientation, the longer sides of the loop ($4.0\text{ cm}$) extend vertically away from the wire rather than running parallel to it. This means that a large portion of the loop’s surface area is pulled significantly further away from the wire where the magnetic field is much weaker. Consequently, both the total magnetic flux passing through the loop and its rate of change over time decrease, resulting in a lower induced emf and a smaller current $I_2$.

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