Electro Magnetic Induction And Alternating Currents Question 364
Question: In a coil of area $ 10\ cm^{2} $ and 10 turns with a magnetic field directed perpendicular to the plane and is changing at the rate of $ 10^{8} $ gauss/second. The resistance of the coil is 20 ohm. The current in the coil will be [CPMT 1976]
Options:
A) 5 amp
B) 0.5 amp
C) 0.05 amp
D) $ 5\times 10^{8}\ amp $
Show Answer
Answer:
Correct Answer: A
Solution:
$ I=\frac{e}{R}=\frac{-N( d\varphi /dt )}{R}=\frac{10\times 10^{8}\times {{10}^{-4}}\times {{10}^{-4}}\times 10}{20} $ = 5 A
BETA
