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



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक