Electro Magnetic Induction And Alternating Currents Question 473

Question: A copper rod of length 0.19 m is moving parallel to a long wire with a uniform velocity of 10 m/s. The long wire carries 5 ampere current and is perpendicular to the rod. The ends of the rod are at distances 0.01 m and 0.2 m from the wire. The emf induced in the rod will be-

Options:

A) 10,μV

B) 20,μV

C) 30,μV

D) 40,μV

Show Answer

Answer:

Correct Answer: C

Solution:

  • EMF induced in an element of length dx at a distance x from wire = Bvdx
    Total EMF induced in the rod E=0.010.2Bvdx=0.010.2μ0iv2πx

    dx=μ0iv2π

    0.010.21x,dx

    E=μ0iv2π[logex]0.010.2=

    μ0iv2π[log10(0.2)log10 (0.01)]×2.303

    E=4π×107×5×102π[1.301]×2.303

    =2.99×105,V30μV



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