Electro Magnetic Induction And Alternating Currents Question 511

Question: An infinitesimally small bar magnet of dipole moment M is pointing and moving with the speed v in the x^ - direction. A small closed circular conducting loop of radius a and negligible self- inductance lies in the y-z plane with its center at x = 0, and its axis coinciding with the x-axis. Find the force opposing the motion of the magnet, if the resistance of the loop is R. Assume that the distance x of the magnet from the center of the loop is much greater than a.

Options:

A) 214μ02M2a4vRx8

B) 163μ0M2a2v2Rx3

C) 323μ0Mav2Rx3

D) None of these

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Answer:

Correct Answer: A

Solution:

  • B=μ02πMx3

    Flux passing through the coil =Ba2

    The induced emf in the coil

    e=dϕdt=ddt[μ0Ma22×1x3]

    =3μ0Ma22x4dxdt[v=dxdt]=3μ0Ma22x4×v

    Current in the coil =I=eR=3μ0Ma2v2x4R

    The magnetic moment of the loop M=I×A=3μ0Ma2v2x4,R×(πa2)

    Now the potential energy of the loop placed in the magnetic field is

    U=MB,cos180o=3μ0Ma2v×πa22x4R×μ0M2πx3

    ,U=3μ02M2a4v4Rx7 Now, |F|=dUdx

    F=214μ02M2a4vRx8

    Since by Newton’s third law, action and reaction are equal.

    Therefore, the above calculated force is acting on the magnet. The direction of the force is in i^ direction by Lenz’s law.



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