Electro Magnetic Induction And Alternating Currents Question 483
Question: A cylindrical region of radius 1 m has instantaneous homogenous magnetic field of 5T and it is increasing at a rate of 2T/s. A regular hexagonal loop ABCDEFA of side 1 m is being drawn in to the region with a constant speed of 1 m/s as . What is the magnitude of emf developed in the loop just after the shown instant when the corner A of the hexagon is coinciding with the centre of the circle?
Options:
A) $ 5/\sqrt{3}V $
B) $ 2\pi /\sqrt{3}V $
C) $ (5\sqrt{3}+2\pi /3),V $
D) $ (5\sqrt{3}+\pi ),V $
Show Answer
Answer:
Correct Answer: C
Solution:
-
The induced emf across the ends B and F due to motion of the loop,
$ e_{1}=Bv(BF)=5\times 1\times 2,\sin ,60^{o}=5\sqrt{3},V $
The induced emf across the loop due to change in magnetic field
$ e_{2}=A\frac{dB}{dt}=\frac{\pi R^{2}}{3}( \frac{dB}{dt} )=\frac{\pi {{(1)}^{2}}}{3}\times 2=\frac{2\pi }{3}V $ So, $ e=e_{1}+e_{2}=( 5\sqrt{3}+\frac{2\pi }{3} )V $
