Question

What is the maximum magnitude of the force on an aluminum rod with a 0.100-μC charge passing between the poles of a 1.50-T permanent magnet at a speed of 5.00 m/s?

A) 7.50 × 10^(-4) N
B) 1.50 × 10^(-3) N
C) 3.75 × 10^(-4) N
D) 6.00 × 10^(-4) N

Answer 1

The maximum magnitude of the force on the aluminum rod is 7.50 x 10^(-4) N.

Step-by-step explanation:

The maximum magnitude of the force on an aluminum rod with a 0.100-μC charge passing between the poles of a 1.50-T permanent magnet at a speed of 5.00 m/s can be calculated using the formula:

F = qvB

where:

F is the force

q is the charge

v is the velocity

B is the magnetic field strength

Substituting the given values, we have:

F = (0.100 x 10^(-6) C) x (5.00 m/s) x (1.50 T) = 7.50 x 10^(-4) N

Therefore, the maximum magnitude of the force is 7.50 x 10^(-4) N.