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 \[B = \sqrt{\frac{mg R sin\theta}{v l^2 \cos^2 \theta}}\]

Answer 1

Component of weight along its motion, F' = mgsinθ

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

\[i = \frac{B \times l \times v cos\theta}{R}\]

\[\left| \overrightarrow{F} \right| = i\left| \overrightarrow{l} \times \overrightarrow{B} \right| = ilB\sin(90 - \theta)\]

\[F = ilB = \frac{Blv \cos\theta}{R} \times l \times B\cos\theta\]

\[F = \frac{B^2 l^2 v \cos^2 \theta}{R}\]

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

\[\frac{B^2 l^2 v \cos^2 \theta}{R} = mg\sin\theta\]

\[\therefore B = \sqrt{\frac{Rmg\sin\theta}{l^2 v \cos^2 \theta}}\]