Question
A loop of wire carrying a current J is initially in the plane of the page and is located in a uniform magnetic field B, which points toward the left side of the page, as shown above. Which of the
following shows the correct initial rotation of the loop due to the force exerted by the magnetic field?
(A) 
(B) 
(C) 
(D) 
(E) 
▶️Answer/Explanation
Ans: A
Use the right-hand rule to determine the force on each part of the loop. A diagram is shown below with the forces indicated. These forces would produce a torque that causes the loop to rotate about the indicated axis. This would cause the loop to rotate as shown in (A).
Question
In the circuit above, what would be the initial current?
(A) \(\frac{44}{5}A\)
(B) \(\frac{215}{62}A\)
(C) \(\frac{25}{22}A\)
(D) \(\frac{62}{215}A\)
(E) \(\frac{5}{44}A\)
▶️Answer/Explanation
Ans:C
First, find the equivalent resistance of the entire circuit.
\(R_{2+3}=R_{2}+R_{3}\)
\(=8\Omega +16\Omega =24\Omega \)
\(\frac{1}{R_{2+3+4}}=\frac{1}{R_{2+3}}+\frac{1}{R_{4}}\)
\(=\frac{1}{24\Omega }+\frac{1}{6\Omega }=\frac{5}{24\Omega }\)
\(R_{2+3+4}=\frac{24}{5}\Omega \)
\(R_{1+2+3+4}=R_{1}+R_{2+3+4}\)
\(=4\Omega +\frac{5}{24}\Omega =\frac{44}{5}\Omega \)
Then use Ohm’s law to solve for current.
V= IR
I = V/R
\((10V)/\left ( \frac{44}{5}\Omega \right )=\frac{25}{22}A\)
Question
Four particles, each with a charge +Q, are held fixed at the corners of a square, as shown above. The distance from each charge to the center of the square is f.
What is the magnitude of the electric field at the center of the square?
(A) $0$
(B) \(\frac{4kQ}{\varphi ^{2}}\)
(C) \(\frac{2kQ}{\varphi ^{2}}\)
(D) \(\frac{4kQ}{\sqrt{2}\varphi ^{2}}\)
(E) \(\frac{kQ}{\sqrt{2}\varphi ^{2}}\)
▶️Answer/Explanation
Ans:A
The strength of each charge’s contribution to the electric field at the center of the square is the same for each charge because they are each the same distance away from the center. The electric field due to each one radiates outward from the charge (not outward from the center). The field created by each is cancelled by the field from the charge diagonal from it, so the total field is zero.
Question
The diagram above shows equipotential lines produced by a charge distribution. A, B, C, D, and E are points in the plane.
At which point is the magnitude of the electric field the greatest?
(A) A
(B) B
(C) C
(D) D
(E) E
▶️Answer/Explanation
Ans: B
The strength of the electric field is given by \(E=-\frac{dV}{dx}\) Therefore, the closer the equipotential lines are the greater the strength of the electric field. B is the point where the equipotential lines have the greatest density, so (B) is correct.