Q. A coil has 1,000 turns and 500 $cm^{2}$as its area. The plane of the coil is placed at right angles to a magnetic induction field of $2 \times 10^{- 5}{Wbm}^{- 2}$. The coil is rotated through 180${^\circ}$in 0.2 s. the average emf induced in the coil, in mV, is

A) 5
B) 10
C) 15
D) 20
Explanation

(b) $N = 1000,\ \ A = 500{cm}^{2} = 500 \times 10^{- 4}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 5 \times 10^{- 2}m^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ B = 2 \times 10^{- 5}\ Wb - m^{- 2},\ \theta_{1} = 0{^\circ},$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \theta_{2} = 180{^\circ},\ \mathrm{\Delta}t = 0.2\ s$
Initial flux linked with coil
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \phi_{1} = NBA\cos\theta_{1}\ $
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = NBA\cos{0{^\circ}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = NBA$
Final flux $\phi_{2} = NBA\cos 180^{0}$
=$\ NBA( - 1) = - NBA$
Change in flux $\phi = \phi_{2} - \phi_{1}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ - NBA - (NBA) = - 2NBA$
$\therefore$ Induced emf
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ e = \frac{- \mathrm{\Delta}\phi}{\mathrm{\Delta}t} = - \frac{( - 2NBA)}{\mathrm{\Delta}t} = \frac{2NBA}{\mathrm{\Delta}t}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ \frac{2 \times 1000 \times 2 \times 10^{- 5} \times 5 \times 10^{- 2}}{0.2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 10 \times 10^{- 3}V = 10\ mV$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.