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:
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 D = $ Bv(\sqrt{2},l)=1\times 1\times 1\times \sqrt{2}=1.41,volt. $
BETA
