Question

A conducting wire $ab$ of length $l$ and resistance $R$, is moving inside a magnetic field $B$ on a pair of fixed vertical rails as shown in the figure. If the mass of the wire is $m$, then calculate the terminal velocity achieved by the loop.

Resistance of the rails are negligible.

• A. $\frac{Rmg}{B^2{l}^2}$
• B. $\frac{3Rmg}{B^2{l}^2}$
• C. $\frac{Rmg}{2B^2{l}^2}$
• D. $\frac{Rmg}{4B^2{l}^2}$

Answer 1

The correct option is A $\frac{Rmg}{B^2{l}^2}$

Due to the gravity, the wire will start to move downwards.

The motion of the wire in magnetic field will induce an emf across the wire.

Let the velocity of wire at any instant is $v$.

$\therefore$ The emf induced across the length of the wire will be,

$E=Bvl$

The induced current is,

$i=\frac{E}{R}=\frac{Bvl}{R}$

Force experienced by the wire due to magnetic filed,

$F=Bil=\frac{B^{2}v{l}^2}{R}$

When this force becomes equal to the weight of the wire, the wire attains terminal velocity.

$\Rightarrow \frac{B^2{v}_t{l}^2}{R}=mg$

$\therefore v_t=\frac{Rmg}{B^2{l}^2}$

Hence, option $\left(A\right)$ is the correct answer.