Search any question & find its solution
Question:
Answered & Verified by Expert
A coil has 1,000 turns and $500 \mathrm{~cm}^2$ as its area. The plane of the coil is placed at right angles to a magnetic induction field of $2 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2$. The coil is rotated through $180^{\circ}$ in 0.2 seconds. The average e.m.f. induced in the coil, in milli-volts, is
Options:
Solution:
2043 Upvotes
Verified Answer
The correct answer is:
$10$
By using $e=-\frac{N B A\left(\cos \theta_2-\cos \theta_1\right)}{\Delta t}$
$e=-\frac{1000 \times 2 \times 10^{-5} \times 500 \times 10^{-4}\left(\cos 180^{\circ}-\cos 0^{\circ}\right)}{0.2}$
$=10^{-2} \mathrm{vol} / \mathrm{t}=10 \mathrm{mV}$
$e=-\frac{1000 \times 2 \times 10^{-5} \times 500 \times 10^{-4}\left(\cos 180^{\circ}-\cos 0^{\circ}\right)}{0.2}$
$=10^{-2} \mathrm{vol} / \mathrm{t}=10 \mathrm{mV}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.