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 due to joule heating is

• 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

Answer: (A)

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, $\epsilon =NAB\omega$
$=N\left(\pi r^2\right)B\omega$
$=20\times \pi \times {\left(6\times 10^{-2}\right)}^2\times 2\times 10^{-2}\times 40$
$=0.18V$
Average value of emf induced over a full cycle ${\epsilon }_{av}=0$
Maximum value of current in the coil,
$I=\frac{\epsilon }{R}=\frac{0.18}{8}=0.023A$
Average power dissipated,
$p=\frac{\epsilon I}{2}=\frac{0.18\times 0.023}{2}$
$=2.07\times 10^{-3}W$