Home / AP Physics C E&M – 13.2 Electromagnetic Induction (Including Faraday’s Law and Lenz’s Law)- Exam Style questions- MCQs
A square wire loop with side \(L\) and resistance \(R\) is held at rest in a uniform magnetic field of magnitude \(B\) directed out of the page, as shown above. The field decreases with time \(t\) according to the equation
\(B=a-bt\), where \(a\) and \(b\) are positive constants.
The current \(I\) induced in the loop is
Correct Answer: \( \boxed{\mathrm{E}} \)
The magnetic flux through the loop is
\(\Phi_B=BA=B(L^2)\)
Since \(B=a-bt\),
\(\dfrac{dB}{dt}=-b\)
Using Faraday’s law, the magnitude of the induced emf is
\(\mathcal{E}=\left|\dfrac{d\Phi_B}{dt}\right|=L^2\left|\dfrac{dB}{dt}\right|=bL^2\)
Applying Ohm’s law, the induced current is
\(I=\dfrac{\mathcal{E}}{R}=\dfrac{bL^2}{R}\)
The magnetic field directed out of the page is decreasing. By Lenz’s law, the induced current must create a magnetic field out of the page to oppose this decrease.
Using the right-hand rule, a magnetic field out of the page is produced by a counterclockwise current.
Therefore, the induced current is \(\dfrac{bL^2}{R}\) and flows counterclockwise.
Therefore, the correct answer is (E).
A magnetic field that is aimed perpendicular to the plane of a wire loop varies as a function of time according to the equation \(B(t)=4t\). Which of the following sketches the emf produced in the loop as a function of time?
Correct Answer: \( \boxed{\mathrm{E}} \)
According to Faraday’s law, the induced emf is
\(\varepsilon=-\dfrac{d\Phi_B}{dt}\)
Since the magnetic field is perpendicular to the loop and the loop’s area remains constant, the magnetic flux is
\(\Phi_B=BA\)
Therefore,
\(\varepsilon=-A\dfrac{dB}{dt}\)
Given
\(B(t)=4t\)
its time derivative is
\(\dfrac{dB}{dt}=4\)
which is a constant. Hence, the magnitude of the induced emf is also constant:
\(\left|\varepsilon\right|=4A\)
Thus, the emf is represented by a horizontal line versus time, corresponding to graph (E).
Therefore, the correct answer is (E).
Two conducting loops that are centered on the same axis carry equal currents \(I\) in the same direction as shown in the diagram above. If the current in the upper loop suddenly decreases to zero, what happens to the current in the lower loop according to Lenz’s law?
Correct Answer: \( \boxed{\mathrm{D}} \)
According to Lenz’s law, the induced current in a circuit always acts to oppose the change in magnetic flux through the circuit.
When the current in the upper loop suddenly decreases to zero, the magnetic field it produces through the lower loop also decreases.
The lower loop responds by producing an induced magnetic field in the same direction as the original field in an attempt to oppose this decrease in magnetic flux.
To produce a magnetic field in the same direction, the induced current in the lower loop must flow in the same direction as its original current.
Therefore, the current in the lower loop increases.
This behavior is consistent with Faraday’s law,
\( \mathcal{E}=-\dfrac{d\Phi_B}{dt}, \)
where the negative sign represents Lenz’s law.
Therefore, the correct answer is (D).