PHYS 1120: General Physics II

Discussion Worksheet 6 Answers

1A. 150 mW
1B. 300 μJ
1C. 1.2 J
1D. 4000 A at 150 mV
1E. 600 V at 1 A

2. segment ab zero force
segment bc 0.04 N in the −x direction
segment cd 0.04 N in the −y direction
segment da 0.04 N in the +x direction + 0.04 N in the +y direction

Notice these add to zero net force, but they do make a nonzero torque

3A. We are given circumference, which is c = 2πr, and asked to find area, which is πr².  We get area A = c²/(4π), which I believe comes out to (1/π)m² with the quantities given here.  Numerically, this comes to 0.3183 m².
3B. Magnetic moment μ = IA = 5.41 mA·m².
3C. The angle is 90 degrees, or π/2 if you are a masochist.
3D. Torque at 90 degrees is μB = 4.33 mN·m.

4A. qA downward (If q is negative, qE downward is upward.)
4B. qvB upward (If q is negative, qvB upward is downward.)
4C. Calling upward positive, ΣF = qvB − qE = q(vB − E)
4D. 0 = q(vB − E) gives v = E/B
4E. The electric force doesn’t depend on speed, but the magnetic force is proportional to speed. If the v > E/B, the particle will deflect up; if v < E/B the particle will deflect down.

5A. qvB
5B. mv²/r = qvB leads to r = mv/(qB).
5C. It all works
5D. Substituting 2πr/T for v eventually gives T = 2πm/(qB)
5E. It all works
5F. Period won’t change, but radius will decrease.  The particle will spiral inward.