Question

A 10-cm-long wire is pulled along a u-shaped conducting rail in a perpendicular magnetic field. the total resistance of the wire and rail is 0.20 ω. pulling the wire at a steady speed of 4.0 m/s causes 4.0 w of power to be dissipated in the circuit.
a. how big is the pulling force?
b. what is the strength of the magnetic field?

Answer 1

In the above case we can say that power given by external agent to pull the rod must be equal to the power dissipated in the form of heat due to magnetic induction.

Part a)

when we pull the rod with constant speed then power required will be product of force and velocity

here we will have

$P = F.v$

P = 4 W

v = 4 m/s

now we will have

$4 = F*4$

$F = 1N$

So external force required will be 1 N

PART B)

now in order to find magnetic field strength we can say

$P = \frac{v^2B^2L^2}{R}$

here we know that induced EMF in the wire is E = vBL

so power due to induced magnetic field is given by

$P = \frac{E^2}{R}$

$4 = \frac{4^2*B^2*0.10^2}{0.20}$

by solving above equation we will have

$B = 2.24 T$