Question

Figure shows a square loop of side $0.5m$ and resistance $10\Omega$. The magnetic field has a magnitude $B=1.0T$. The work done in pulling the loop out of the field slowly and uniformly in $2.0s$ is

• A. $3.125\times {10}^{-3}J$
• B. $6.25\times {10}^{-4}J$
• C. $1.2\times {10}^{-2}J$
• D. $5.0\times {10}^{-4}J$
• E. answer required

Answer 1

The correct option is A $3.125\times {10}^{-3}J$

$emf=Bl\frac{l}{2}$
work done in pulling out = $\frac{em{f}^2}{R}\times t$
$=\frac{B^2{l}^2\frac{l^2}{2^2}}{R}\times 2=3.125mJ$