Question

A circular coil of radius $6\, cm$ and $20$ turns rotates about its vertical diameter with an angular speed of $40\, rad \, s^{-1}$ 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

The Correct Option is A

Here, $r = 6 cm = 6 \times 10^{-2}\,m$, $N=20, \omega=40\,rad\,s^{-1}$ $B=2\times10^{-2}\,T, R=8\, \Omega$ Maximum emf induced, $\varepsilon=NAB\omega$ $=N\left(\pi r^{2}\right)B\omega$ $=20\times\pi\times\left(6\times10^{-2}\right)^{2}\times2\times10^{-2}\times40$$=0.18\,V $ Average value of emf induced over a full cycle $\varepsilon_{av}=0$ Maximum value of current in the coil, $I=\frac{\varepsilon}{R}=\frac{0.18}{8}=0.023\,A$ Average power dissipated, $p=\frac{\varepsilon I}{2}=\frac{0.18\times0.023}{2}$ $=2.07\times10^{-3}\,W $