Electro Magnetic Induction And Alternating Currents Question 509
Question: A square metal wire loop of side 10 cm and resistance 1 ohm is moved with a constant velocity $ v_{0} $ in a uniform magnetic field of induction $ B=2,weber/m^{2} $ as . The magnetic field lines are perpendicular to the plane of the loop (directed into the paper). The loop is connected to a network of resistors each of value 3 ohms. The resistances of the lead wires OS and PQ are negligible. What should be the speed of the loop so as to have a steady current of 1 milliampere in the loop? Give the direction of current in the loop.
Options:
A) 2.3 m/s
B) 1.2 m/s
C) 0.3 m/s
D) 0.02 m/s
Show Answer
Answer:
Correct Answer: D
Solution:
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The network behaves like a balanced wheatstone bridge.
The induced emf developed is given by $ e=vB\ell =v\times 2\times 0.1=0.2v $ …(i)
Now, $ e=IR $ $ e={{10}^{-3}}\times 4=4\times {{10}^{-3}}amp $ …(ii) From (i) and (ii), $ 0.2,v=4\times {{10}^{-3}} $
$ \therefore ,v=\frac{4\times {{10}^{-3}}}{0.2}=0.02m/s $
BETA
