Physics - EMI/AC Question with Solution | TestHub

Speed of the loop should be

$v=\frac{\mathrm{length}}{\mathrm{time}}$$=\frac{\text{0.5}}{2}={\text{0.25 m s}}^{-1}$

Induced emf$\text{e}=\mathit{\text{Bv}}l=\left(\text{1.0}\right)\left(\text{0.25}\right)\left(\text{0.5}\right)=\text{0.125 V}$

$\therefore$Current in the loop$i=\frac{e}{R}=\frac{\text{0.125}}{10}=\text{1.25}\times 10^{-2}\text{ A}$

The magnetic force on the left arm due to the magnetic field is

$F_{\text{m}}=ilB=\left(\text{1.25}\times 10^{-2}\right)\left(\text{0.5}\right)\left(\text{1.0}\right)$

$=\text{6.25}\times 10^{-3} N$

To pull loop uniformly an external force of$\text{6.25}\times {10}^{-3} N$towards right must be applied.

$\therefore$$W = (6.25 \times {10}^{-3} \mathrm{N}) (0.5 \mathrm{m})$

$= 3.125 \times {10}^{-3} \mathrm{J}$