Mock Exams AP Physics – C : E & M– MCQ Set 1

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.

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