Question

A square loop of area $25\, cm ^2$ has a resistance of $10 \Omega$. The loop is placed in uniform magnetic field of magnitude $40.0 \,T$. 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 $1.0 \sec$, will be

• A. $2.5 \times 10^{-3} J$
• B. $1.0 \times 10^{-3} J$
• C. $1.0 \times 10^{-4} J$
• D. $5 \times 10^{-3} J$

Answer 1

Answer: (B)

Sol. $\ell=50\, cm$
$ t =1 \sec$
$\therefore V =\frac{0.05}{1}=0.05\, m / s . $
$ i =\frac{40 \times 0.05 \times 0.05}{10}=0.01 A$
$\therefore F = B _{ i } \ell=40 \times 0.01 \times 0.05$
$F =0.02 \,N $
$\therefore W =0.02 \times \ell=0.02 \times .05 $
$\therefore W =1 \times 10^{-3} J$