AP Physics 2- 12.1 Magnetic Fields- Exam Style questions - FRQs- New Syllabus

(a)
Since the charged object moves with constant velocity, the net force on the object is \(0\). Therefore, the magnetic force and electric force must have equal magnitudes.

\(F_M=F_E\)

The electric force on the charged object has magnitude \(F_E=qE\), where \(q\) is the magnitude of the charge.

The magnetic force on the moving charged object has magnitude \(F_M=qvB\), because the velocity is perpendicular to the magnetic field at the location of the object.

Set the magnitudes equal:

\(qvB=qE\)

Cancel \(q\):

\(vB=E\)

The magnetic field a distance \(d\) from a long, straight current-carrying wire is

\(B=\dfrac{\mu_0 I}{2\pi d}\)

Substitute this expression into \(vB=E\):

\(v\left(\dfrac{\mu_0 I}{2\pi d}\right)=E\)

Solve for \(v\):

\(\boxed{v=\dfrac{2\pi dE}{\mu_0 I}}\)

The charge \(q\) cancels because both the electric and magnetic forces are proportional to the charge magnitude.

(b)(i)
The “\(X\)” should be placed between point \(P_1\) and the dashed horizontal line through the middle of the coil.

The magnetic field due to a long, straight wire is inversely proportional to the distance from the wire:

\(B=\dfrac{\mu_0 I}{2\pi r}\)

Since \(B\propto \dfrac{1}{r}\), the magnetic field is stronger closer to the wire and weaker farther from the wire.

At point \(P_1\), the magnetic field is \(3B_0\). At point \(P_2\), the magnetic field is \(B_0\). A field of \(2B_0\) is between these values, so the point must be somewhere between \(P_1\) and \(P_2\).

More specifically, because \(P_1\) is closer to the wire and \(2B_0\) is less than \(3B_0\), the \(X\) must be above \(P_1\). Because \(2B_0\) is still greater than \(B_0\), it must be below \(P_2\). The correct location is between \(P_1\) and the dashed line.

(b)(ii)
Use Faraday’s law to find the magnitude of the induced emf:

\(|\varepsilon|=\left|\dfrac{\Delta \Phi_B}{\Delta t}\right|\)

The change in magnetic flux is

\(\Delta \Phi_B=\Phi_{B,f}-\Phi_{B,i}\)

\(\Delta \Phi_B=\left(1.0\times 10^{-5}\,\text{T}\cdot\text{m}^2\right)-\left(5.0\times 10^{-5}\,\text{T}\cdot\text{m}^2\right)\)

\(\Delta \Phi_B=-4.0\times 10^{-5}\,\text{T}\cdot\text{m}^2\)

The magnitude is

\(|\Delta \Phi_B|=4.0\times 10^{-5}\,\text{T}\cdot\text{m}^2\)

So the magnitude of the induced emf is

\(|\varepsilon|=\dfrac{4.0\times 10^{-5}\,\text{T}\cdot\text{m}^2}{2.0\,\text{s}}\)

\(|\varepsilon|=2.0\times 10^{-5}\,\text{V}\)

Now use Ohm’s law:

\(I_{\text{coil}}=\dfrac{|\varepsilon|}{R}\)

\(I_{\text{coil}}=\dfrac{2.0\times 10^{-5}\,\text{V}}{10\,\Omega}\)

\(\boxed{I_{\text{coil}}=2.0\times 10^{-6}\,\text{A}}\)

The average current is very small, but it is nonzero because the magnetic flux through the coil is changing.

(c)
The current in the round coil produces a magnetic field around the round coil.

Since the square coil is placed above and concentric with the round coil, some of the magnetic field from the round coil passes through the area enclosed by the square coil. This produces magnetic flux through the square coil.

During the time interval, the current in the power supply is constantly increasing. Since the current in the round coil increases, the magnetic field produced by the round coil also increases.

Because the magnetic field through the square coil changes, the magnetic flux through the square coil changes.

By Faraday’s law, a changing magnetic flux induces an emf:

\(|\varepsilon|=\left|\dfrac{\Delta \Phi_B}{\Delta t}\right|\)

The induced emf causes an induced current in the square coil circuit. Since the lightbulb is connected in that circuit, the induced current passes through the lightbulb.

Therefore, the lightbulb remains lit during the entire time interval because the current in the round coil is changing during the entire time interval, so the magnetic flux through the square coil is also changing during the entire time interval.

\(\boxed{\text{Changing current } \rightarrow \text{ changing magnetic field } \rightarrow \text{ changing flux } \rightarrow \text{ induced current in the bulb.}}\)