Question

A uniform magnetic field is directed into the screen. There are two parallel conducting rails, running horizontally, in the field, with a conducting rod on top of the rails. The rails are a distance L apart. The picture shows a force F directed to the right on the rod. The rails are joined at the left by a resistor of resistance R.

​We'll use these values:

L = 20 cm; B = 4.0 T; F = 3.2 N; and R = 1.0 ohms.

1. The rod starts from rest, and slides on the rails without friction. Even though the force F continues to be applied, the rod eventually reaches a constant velocity to the right. What is the constant speed reached by the rod?
_____ m/s

2. When the rod is going at constant velocity, what is the magnitude of the induced current in the loop consisting of the rod, rails, and resistor?
_____ A

3. Let's say that you are applying the force to the right. The power you input to the system to move the rod ends up being dissipated in the resistor. Calculate this power.

Answer 1

1) induced emf in the loop, emf = B*v*L

induced current, I = emf/R

= B*v*L/R

when the rod moves with constant speed,

applied force = magnetic force in the opposite direction

F_applied = B*I*L

F_applied = B*( B*v*L/R)*L

F_applied = B^2*L^2*v/R

==> v = F_applied*R/(B^2*L^2)

= 3.2*1/(4^2*0.2^2)

= 5.0 m/s <<<<<<<<<--------------Answer

2) induced current, I = emf/R

= B*v*L/R

= 4*5*0.2/1

= 4.0 A <<<<<<<<<--------------Answer

3) Power dissipated, P = I^2*R

= 4^2*1

= 16 W <<<<<<<<<--------------Answer