Electro Magnetic Induction And Alternating Currents Question 410

Question: A metal conductor of length 1 m rotates vertically about one of its ends at an angular velocity 5 radians per second. If the horizontal component of the earth’s magnetic field is $ 0.2\times {{10}^{-4}}T $ , then the emf developed between the two ends of the conductor is

Options:

A) $ 5,\mu V $

B) $ 50,\mu V $

C) $ 5,mV $

D) $ 50,mV $

Show Answer

Answer:

Correct Answer: B

Solution:

  • $ E=\frac{1}{2}B\omega R^{2}=\frac{1}{2}\times 0.2\times {{10}^{-4}}\times 5\times 1\times 1 $

    $ =0.5\times {{10}^{-4}}V=54\times {{10}^{-1}}V $

    $ =50\times {{10}^{-6}}V=50\mu V $



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक