Question

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

• A. $3.125\times 10^{- 3}J$
• B. $6.25\times 10^{- 4}J$
• C. $1.25\times 10^{- 2}J$
• D. $5.0\times 10^{- 4}J$

Answer 1

Answer: (A)

Speed of the loop should be
$v=\frac{length}{time}$ $=\frac{\text{0.5}}{2}=\text{0.25 m s}^{- 1}$
Induced emf $\text{e}=\textit{Bv}l =\left(\text{1.0}\right)\left(\text{0.25}\right)\left(\text{0.5}\right)=\text{0.125 V}$
$\therefore $ Current in the loop $i=\frac{e}{R}=\frac{\text{0.125}}{1 0}=\text{1.25}\times 10^{- 2}\text{ A}$
The magnetic force on the left arm due to the magnetic field is
$F_{\text{m}}=ilB=\left(\text{1.25} \times 1 0^{- 2}\right)\left(\text{0.5}\right)\left(\text{1.0}\right)$
$=\text{6.25}\times 10^{- 3}N$
To pull loop uniformly an external force of $\text{6.25}\times 10^{- 3}N$ towards right must be applied.
$\therefore $ $W=\left(\right.6.25\times 10^{- 3}N\left.\right)\left(\right.0.5m\left.\right)$
$=3.125\times 10^{- 3}J$