Physics Problems & Solutions: Electromagnetism

Two small pith balls, each carrying a charge q, are attached to the ends of a light rod of length d, which is suspended from the ceiling by a thin torsion-free fiber, as shown in the figure. There is a uniform magnetic field B, pointing straight down, in the cylindrical region of radius R around the fiber. The system is initially at rest. If the magnetic field is turned off, which of the following describes what happens to the system?

It rotates with angular momentum qBR².
It rotates with angular momentum ¼ qBd².
It rotates with angular momentum ½ qBRd.
It does not rotate because to do so would violate conservation of angular momentum.
It does not move because magnetic forces do no work.

(GR9677 #87)

Solution:

B and C are FALSE.
Angular momentum is finite, it cannot go to infinite.
If d → ∞ , angular momentum → ∞

D. FALSE
Since there is external torque, angular momentum is not conserved.

E. FALSE
The system will rotate due to B.

Answer: A:

Calculation:
Faraday's Law: $\varepsilon=-\frac{d \Phi _B}{dt}=\vec{E}\cdot dl$

with Φ = BπR² and dl = 2πd/2 = πd

$\pi R^2 \frac{dB}{dt}=E\pi d\rightarrow E=\frac{\dot{B}R^2}{d}$
$F=qE=\frac{q\dot{B}R^2}{d}$

Torque: $\tau =\sum\vec{r}\times \vec{F}$

Since there is a force contribution from each charge and by the right-hand-rule their cross products with the moment-arm point in the same direction,

$\tau =2 \frac{d}{2}\frac{q\dot{B}R^2}{d}=q\dot{B}R^2$

Torque and Angular momentum:

$\\\sum \tau = \dot{L} \\\rightarrow \dot{L} = q\dot{B}R^2 \\\rightarrow L= qBR^2$

Physics Problems & Solutions

Electromagnetism - Angular Momentum

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