A square loop of area $25 \mathrm{~cm}^2$ has a resistance of $10 \Omega$. The loop is placed in uniform…
A square loop of area $25 \mathrm{~cm}^2$ has a resistance of $10 \Omega$. The loop is placed in uniform magnetic field of magnitude 40 T . The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in 1 second, will be
$\quad 2.5 \times 10^{-3} \mathrm{~J}$
$1.0 \times 10^{-3} \mathrm{~J}$
$1.0 \times 10^{-4} \mathrm{~J}$
$5 \times 10^{-3} \mathrm{~J}$
Solution
Area of square is $25 \mathrm{~cm}^2$
$\therefore \quad$ length $=5 \mathrm{~cm}=0.05 \mathrm{~m}$
Velocity, $\mathrm{v}=\frac{l}{\mathrm{t}}=\frac{0.05}{1}=0.05 \mathrm{~m} / \mathrm{s}$
$\mathrm{e}=\mathrm{B} / \mathrm{v} \quad \Rightarrow \mathrm{IR}=\mathrm{B} / \mathrm{v}$
$\therefore \quad \mathrm{I}=\frac{\mathrm{B} / \mathrm{v}}{\mathrm{R}}=\frac{40 \times 0.05 \times 0.05}{10}=0.01 \mathrm{~A}$
Since magnetic field is perpendicular to plane of loop, $\theta=90^{\circ}$
$\begin{aligned}
\therefore \quad \mathrm{F} & =\mathrm{BI} l \sin \theta \\
& =40 \times 0.01 \times 0.05 \times \sin 90=0.02 \mathrm{~N} \\
\therefore \quad \mathrm{~W} & =\mathrm{F} \times l \\
& =0.02 \times 0.05=1 \times 10^{-3} \mathrm{~J}
\end{aligned}$
.