Download PHYS 212 Examination 3: Problems and Solutions in Electromagnetism - Prof. M. Hackwrth and more Exams Physics in PDF only on Docsity!
PHYS 212 Examination 3
Name (print): _____________________________
Signature _____________________________
Problem 1 __________________
Problem 2 __________________
Problem 3 __________________
Problem 4 __________________
Problem 5 __________________
Total __________________
Directions: This exam contains five problems worth 20 points each for a possible 100
points. Your solutions should be written as neatly as possible and arranged in a logical
manner. Credit will be awarded on the basis of thought, compactness, and neatness of the
written solution. Remember to use basic physical principles in solving the problems. Show
all of your work. I will not award full points for a problem with a solution that I am unable to
decipher even if the answer is correct.
An equation sheet has been provided. CRC handbooks are allowed. Calculator rule is in
effect. Good Luck!
Problem 1.
- A proton is released from rest at the origin of the coordinate system shown below. A
uniform E
r
field exists in the + j
direction. A uniform B
r
field exists in the + i
direction.
Sketch the path of this particle.
Cycloid along k
− axis
- A proton enters a magnetic field B = 3 i – 2 j + k with a velocity of v = 3 i - 2 j - k. What is
the magnitude and direction of the resulting force on the proton?
× = − − × − + = − −
i j k
v B i j k i j k
v r
i
j
( i j k ) ( k i j )
i j F [ q ( i j )] N
r
B
E
Problem 3. Determine the current that flows through the 100 Ω resistor in the circuit below.
221mA
Problem 4. A rectangular loop consists of 10 turns and has the dimensions given below. The
loop is hinged on the y -axis, and its plane makes an angle of 30
0
with the x -axis. A uniform
magnetic field B = 0.8 i T exists in the region of space containing the loop and a current of 1.
amperes flows through the loop in the indicated direction.
- What is the net force on the loop?
- What is the magnetic moment of the loop?
- What is the torque on the loop?
- What is the angular acceleration of the loop?
- The net force is zero.
- The magnetic moment, IA , is 0.144 a/m
2
and is perpendicular to the plane of the loop in
the direction of the x-y plane.
- Γ = n ( m B sin α) ≈ 1 N ⋅ m
r
(α = 60
0
). The torque vector points in the direction of n B
r
×
( j
I
I , so the angular acceleration is the torque divided by the moment of inertia
and is in the j
− direction.
