Electro Magnetic Induction And Alternating Currents Question 334

Question: A conductor ABOCD moves along its bisector with a velocity of 1 m/s through a perpendicular magnetic field of 1 wb/m2, as shown in fig. If all the four sides are of 1m length each, then the induced emf between points A and D is

Options:

A) 0

B) 1.41 volt

C) 0.71 volt

D) None of the above

Show Answer

Answer:

Correct Answer: B

Solution:

There is no induced emf in the part AB and CD because they are moving along their length while emf induced between B and C i.e. between A and D can be calculated as follows Induced emf between B and C = Induced emf between A and B = $ Bv(\sqrt{2},l)=1\times 1\times 1\times \sqrt{2}=1.41,volt. $



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