Question

A rectangular frame of wire $abcd$ has dimensions $32cm\times 8.0cm$ and a total resistance of $2.0\Omega$. It is pulled out of a magnetic field $B=0.020T$ by applying a force of $3.2\times {10}^{-5}N$ (figure). It is found that the frame moves with constant speed. What is constant speed of the frame?

Answer 1

If loop abcd moves with constant speed ,therefore electromagnetic force $f_m$ is equal to external force $f$.

$f_m=f=3.2\times {10}^{-5}$

$ilB=3.2\times {10}^{-5}$

$\frac{e}{R}lB=\frac{VBl}{R}lB=\frac{V{B}^2{l}^2}{R}$ where,

$V$=speed of loop

$R$=resistance of abcd.

$i$=current in the loop.

$B$=magnetic field

$⟹\frac{V\times {\left(0.2\right)}^2\times {(8\times {10}^{-2})}^2}{2}=3.2\times {10}^{-5}$

$⟹\frac{V\times 4\times {10}^{-4}\times 64\times {10}^{-4}}{2}=3.2\times {10}^{-5}$

$⟹V=\frac{2\times 3.2\times {10}^3}{4\times 64}=25\frac{m}{s}$

Velocity required =$25m{s}^{-1}$