Question

A circular coil of radius $6cm$ and $20$ turns rotates about its vertical diameter with an angular speed of $40rad{s}^{-1}$ in a uniform horizontal magnetic field of magnitude $2\times {10}^{-2}$ T. If the coil form a closed loop of resistance 8 $\Omega$, then the average power loss

• A. $2.07\times {10}^{-3}$W
• B. $1.23\times {10}^{-3}$W
• C. $3.14\times {10}^{-3}$W
• D. $1.80\times {10}^{-3}$W

Answer 1

The correct option is A $2.07\times {10}^{-3}$W

Here, $r=6cm=6\times {10}^{-2}m,$

$N=20,\omega =40rad{s}^{-1}$

$B=2\times {10}^{-2}T,R=8\Omega$

Maximum emf induced, $ϵ=NAB\omega$

$=N\left(\pi r^2\right)B\omega$

$=20\times \pi \times (6\times {10}^{-2}{)}^2\times 2\times {10}^{-2}\times 40=0.18V$

Average value of emf induced over a full cycle ${ϵ}_{av}=0$

Maximum value of current in the coil, $I=\frac{\epsilon }{R}=\frac{0.18}{8}=0.023A$

Average power dissipated, $P=\frac{\epsilon I}{2}=\frac{0.18\times 0.023}{2}=2.07\times {10}^{-3}W$