Question

A wire ab of length l, mass m and resistance R slides on a smooth, thick pair of metallic rails joined at the bottom as shown in figure. The plane of the rails makes an angle θ with the horizontal. A vertical magnetic field B exists in the region. If the wire slides on the rails at a constant speed v, show that B=√mg R sinθvl2cos2 θ
Figure

Answer 1

EMF is Product of Length, Velocity and Field

Explanation:

Component of weight along with its motion, $F^'$ = $mgsin\theta$

The emf induced in the rod due to its motion is given by

e = Bl'v'

Here,

$l^'$ = Component of the length of the rod perpendicular to the magnetic field

$v^'$ = Component of the velocity of the rod perpendicular to the magnetic field

i = $\frac {B\times \timesl \timesv cos\theta}{R}$

$\left | \vec{F} \right | = i\left | \vec{l}\vec{B} \right | = ilBsin (90 - \theta)$

$F = ilB = \frac {B^2l^2vcos^2\theta}{R}$  

The direction of force F is opposite to $F^'$

• Because the rod is moving with a constant velocity, the net force on it is zero.

Thus,

F - $F^'$ = 0

F = $F^'$

or

$\frac {B^2l^2vcos^2\theta}{R} = mgsin\theta$

B = $\sqrt (\frac {vRmg sin\theta}{l^2vcos^2\theta})$