Question

A wire $\text{CD}$ of length $l=0.5\text{m}$ and mass $m=1\text{kg}$ is sliding without friction on conducting rails as shown. The vertical rails are connected together with a resistance $R=4\Omega$. A uniform magnetic field $B=1\text{T}$ is applied perpendicular to the plane of wires such that $\text{CD}$ moves with a constant velocity $v$. Calculate $v\left(\text{in m/s}\right)$.
(Give integer value only)

Answer 1

Emf induced in the circuit is,

$E=Bvl=1\times v\times 0.5=0.5v$

$\therefore$ Current flowing in the circuit is,

$i=\frac{E}{R}=\frac{0.5v}{4}$

$\therefore$ Force acting on wire due to magnetic field,

$F_1=ilB$

$=\left(\frac{0.5v}{4}\right)\times 0.5\times 1=\frac{2.5v\times {10}^{-1}}{4}\text{N}$

Since, wire is moving with constant speed, therefore, magnetic force acting on the wire is equal to the force of gravity.

$\therefore mg=ilB$

$\Rightarrow \frac{1}{2}\times 10=\frac{2.5\times {10}^{-1}v}{4}$

$\Rightarrow v=80\text{m/s}$

Correct answer: $80$