# Midterm Exam 2 Information

The second midterm exam will take place at 10AM on Tuesday April 8th. You will have 80 minutes to complete the exam. Make sure to be on time as the exam will finish promptly at 11:20AM.

You may bring two sheets of letter size paper which have handwritten notes on both sides. You should also bring a calculator.

The exam will consist of 3 questions. The breakdown of questions will be approximately one question on electromagnetic fields and forces, one question on sources of magnetic fields and one question on induction. Elements of electrostatics may be included in any or all of the questions.

## Results

The average on the midterm was 79.5%. Here is a histogram of the scores.

## Question 1 Solution (Average Score: 26.8/30 (89.3%))

A. $B=\frac{\mu_{0}I}{2r}\times\frac{1}{4}=\frac{\mu_{0}I}{8r}$ in to page

B. $B=0\mathrm{T}$

C. $B=\frac{\mu_{0}I}{8r}-\frac{3\mu_{0}I}{16r}-\frac{3\mu_{0}I}{32r}=\frac{5\mu_{0}I}{32r}$ out of page.

## Question 2 Solution (Average Score: 27.8/35 (79.3%) )

A. $B=\frac{\mu_{0}I}{2\pi r}=\frac{2\times10^{-5}}{r}$

B. $3.33\times10^{-4}\mathrm{T}$ out of the page.

C. The magnitude of the field at $y$ is smaller than the field at $x$.

D. In fact the correct answer is the tension is simply equal to $mg=9.8\times5\times10^{-3}=0.049\mathrm{N}$.

However you could also get full points if you correctly calculated the magnetic field on the bent wire.

$F_{mag}=2\frac{\mu_{0}I_{1}I_{2}}{2\pi}\times\frac{1}{2}\int_{6\mathrm{cm}}^{11.196\mathrm{cm}}\frac{dy}{y}=\frac{4\pi\times10^{-7}\times100\times70}{2\pi}\ln{\frac{11.196}{6}}=0.00087\mathrm{N}$

E. A clockwise current should flow in the weight which increases until the weight passes the wire, after which a diminishing anticlockwise current would be seen.

## Question 3 Solution (Average Score: 24.9/35 (71.1%))

A. $2Bl=\mu_{0}lj_{0}d$

$\vec{B}=\frac{\mu_{0}j_{0}d}{2}\hat{i}$

B. $0\mathrm{A}$

C. $\mathcal{E}=-A\frac{dB}{dt}=-\pi r^{2}\frac{\mu_{0}j_{0}d}{2\tau}$

$I=\frac{\mathcal{E}}{R}=\frac{\pi r^{2}\mu_{0}j_{0}d}{2\tau R}$

D. $B_{ring}=\frac{\mu_{0}I}{2r}$

$B_{total}=[\frac{\mu_{0}^{2}\pi r j_{0}d}{4\tau R}+\frac{\mu_{0}j_{0}d}{2}(1-\frac{t}{\tau})]\hat{i}$

E. $W=Pt=\mathcal{E}I\tau=\frac{\pi^{2}r^{4}\mu_{0}^{2}j_{0}^{2}d^{2}}{4R\tau}$

F. No, the energy is not recovered, even though the current flows in the opposite direction when the power is turned back on, energy is lost in both processes as energy is lost each time current flows through a resistor, independent of direction.

## Practice Exams

There are two practice exams posted on Mastering Physics. There are also several extra practice problems, also on Mastering Physics. Below you can find midterms from previous years and solutions to them.