PHY 132 Second Midterm, 4/2001, L. Mihaly

Remove everything from work area, except this exam, your one page of handwritten notes and your calculator. Wait for instruction to start. Use the paper provided for all work (back side is OK), ask for empty sheets if you need more. When finished, copy all results to this (front) page. Hand in all your work.

Leaving the exam room: Ask for permission. Do not talk to anyone outside. Do not remove the exam sheet from the room until the exam is over. If finished early, remain seated until 10 minutes before the exam ends. Leave the room during the last 10 minutes ONLY if you are finished and do not want to return.

Scores: Problems 1,2,3: 10 points for correct answer, partial credit possible. Problems 4-7: 5 points each, no partial credit.

Complaints: If you feel that a mistake was made, return the exam and a brief written statement of your grievance to your recitation instructor. Preferably, you should ask for the re-grading at the end of the recitation session. Except for trivial numerical mistakes (like incorrect addition of the scores by the grader), the returned exam book will be accepted only on the same day that you received the graded exam in recitation. Do not change your solution in any way. If you annotate the exam sheet, use a writing instrument that is markedly different from the pen/pencil used during the exam. If partial credit is contested, the whole exam will be re-graded. Your grade may go up, down or remain unchanged.

1. What is the current through the R₃ resistor shown in the Figure. The batteries have emfs of E₁=9.0V and E₂=12.0V, and the resistances have values of R₁=15W, R₂=20W , and R₃ = 40W. In your response, specify the magnitude and the direction (up or down).

Use Kirchoff's laws. Notice that the polarity of the batteries is opposite. Take I₁ to the right, I₂ to the left, I₃ upwards. For the upper loop: 9-15I₁-20I₂ =0. for the lower loop: 12-40I₃-20I₂ =0. The junction law tells us I₂=I₁+I₃. Solve for I₃ by eliminating I₂ first, and then I₁. The result is I₃= 0.141A, up.

2. A proton moving with speed v=4.0x10⁵m/s in a field-free region abruptly enters an essentially uniform magnetic field B=0.850T as shown in the Figure. The velocity is perpendicular to the field. If the proton enters the magnetic field region at a 45^o angle as shown, then the angle q at the exit point is also 45^o. At what distance x does it leave from the field? (The mass of the proton is 1.27x10^-27kg, the magnitude of the charge is the same as the electron charge.)

The proton moves on a circular path. Geometry tells us that the distance x satisfies x/2= rcos45^o, where r=mv/qB is the radius of the cyclotron orbit. We get x= 1.41mv/qB= (1.41)(1.27x10^-27kg)(4x10⁵m/s)/[(1.6x10^-19C)(0.85T)]=6.92x10^-3m=5.27mm

3. A long straight pair of wires serves to conduct 50.0A current to and from and instrument. The wires are of negligible diameter, and they run parallel, 3.4cm apart. What is the magnitude of the magnetic field 10cm from the center line of the wires, in the plane of the wires?

Let us use r₀=10cm, d=3.4cm. The field of one wire B₁= m₀/(2p)(I/r₁), where r₁= r₀+d/2. The other one creates an opposite field, with r₂= r₀-d/2. The magnitude of the total field is B= Im₀/(2p)(1/r₁-1/r₂)= Im₀/(2p)d/(r₀²-d²/4)=3.5x10^-5T .

4. A 75W 120V light bulb is in parallel with a 40W 120V light bulb. What is the net resistance?

A 1.04W

B 4.60W

C 125W

D 192W

E 552W

F none of the above

The total power is 115W. From P=V²/R, we get R= P=V²/P= (120V)²/(115W)= 125W (221W). Another way of solving this is by calculating R₁=(110V)²/(75W), R₂=(110V)²/(40W), and using 1/R=1/R₁+1/R₂.

5. A 50.0cm x 12.0cm rectangular coil with 200 turns carries a current of 8.3A. What is the maximum torque on the coil if it is in a uniform magnetic field of 0.300T?

A 4.48x10⁵Nm

B 3.00x10⁵Nm

C 44.8Nm (N=300)

D 30.0Nm (N=200)

E 0.15Nm

F none of the above

The largest torque acts when the field is parallel to the plane of the square. The magnitude of the torque is t = NABI = 30.0Nm .

6. A rod of length L=0.50m slides on a pair of rails in a magnetic field of 0.500T. The end of the rails is connected to a resistor of 10W (see Figure). The magnetic field is perpendicular to the rod. The velocity of the rod is 8.5m/s. What is the current in the resistor?

A 0.21A (8.5m/s)

B 0.16A (6.5m/s)

C 2.1A

D 1.6A

E 3.7A

F none of the above

The induced EMF is BvL, and this creates a current of magnitude I=BvL /R=0.21A.

7. What is the current in a 0.036mF capacitor connected to a 120V, 60Hz line?

A 0.259mA

B 1.63mA

C 0.259A

D 1.63A

E 3.28A

F none of the above

The impedance is X_C=1/(wC)=1/(2pfC)= 74kW. The current is 120V/74kW =1.63mA