Question

a length of thin copper wire is used to wind a square coil that is l=25cm on each side. the coil has a single turn.

Answer 1

Area vector and magnetic field are in same direction hence initial angle will be 0

Torque acting on the coil is given as

$\tau =NIAB\sin\theta$

Current in the loop is

$I=\frac{V}{R}=\frac{9}{0.02}=450A$

Areal of loop is

$A=L^2=0.25\times0.25=0.0625m^2$

Work done in rotating the loop is given as

$W=\int_{\theta_1}^{\theta_2}NBAI\sin\theta d\theta=-NBAI[\cos\theta]_{0}^{50}=NBIA[\cos0-\cos50]\\ \Rightarrow W=1\times2\times0.0625\times450\times[1-0.64287]=20.093\approx 20.1J$