AP Physics C- E&M- 12.3 Magnetic Fields of Current-Carrying Wires - Exam Style questions - FRQs- New Syllabus

A.
\(\boxed{F_2>F_1}\)

The magnetic force on a moving charged particle is

\(F=qvB\sin\theta\)

Here, the spheres move in the \(+x\)-direction, while the magnetic fields from the long wires are in the \(+z\) or \(-z\) direction. Therefore, \(\theta=90^\circ\), so

\(F=QvB\)

Thus, the sphere with the larger net magnetic field experiences the larger magnetic force.

At Sphere \(1\), the magnetic fields from wires \(S\) and \(T\) are in opposite directions. Sphere \(1\) is a distance \(d\) from wire \(S\) and \(3d\) from wire \(T\), so the field from wire \(S\) is larger than the field from wire \(T\). The fields partially cancel.

At Sphere \(2\), the sphere is a distance \(d\) from each wire. The magnetic fields from wires \(S\) and \(T\) are in the same direction, so they add.

Therefore, the net magnetic field at Sphere \(2\) is larger than the net magnetic field at Sphere \(1\), so

\(\boxed{F_2>F_1}\)

B.
For a long straight current-carrying wire, the magnetic field at distance \(r\) from the wire is

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

Sphere \(2\) is a distance \(d\) from wire \(S\) and a distance \(d\) from wire \(T\).

The magnetic field due to wire \(S\) at Sphere \(2\) is

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

The magnetic field due to wire \(T\) at Sphere \(2\) is also

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

Using the right-hand rule, the magnetic fields from the two wires at Sphere \(2\) point in the same direction, so the magnitudes add:

\(B_{\text{tot}}=B_S+B_T\)

\(B_{\text{tot}}=\dfrac{\mu_0 I}{2\pi d}+\dfrac{\mu_0 I}{2\pi d}\)

\(\boxed{B_{\text{tot}}=\dfrac{\mu_0 I}{\pi d}}\)

C.
\(\boxed{F_{\text{new}}=F_2}\)

From part B, the magnetic force magnitude is proportional to the net magnetic field magnitude:

\(F=QvB_{\text{tot}}\)

In the new situation, wire \(S\) still carries current \(I\), so its field at Sphere \(2\) has magnitude

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

Wire \(T\) now carries current \(3I\), so its field at Sphere \(2\) has magnitude

\(B_T=\dfrac{\mu_0(3I)}{2\pi d}=\dfrac{3\mu_0 I}{2\pi d}\)

In Figure \(2\), the fields from wires \(S\) and \(T\) at Sphere \(2\) are in opposite directions, so their magnitudes subtract:

\(B_{\text{new}}=\dfrac{3\mu_0 I}{2\pi d}-\dfrac{\mu_0 I}{2\pi d}\)

\(B_{\text{new}}=\dfrac{2\mu_0 I}{2\pi d}\)

\(B_{\text{new}}=\dfrac{\mu_0 I}{\pi d}\)

This is the same value found in part B:

\(B_{\text{new}}=B_{\text{tot}}\)

Since \(Q\) and \(v\) are unchanged,

\(\boxed{F_{\text{new}}=F_2}\)