Electromagnetic Induction and Alternating Current 7 Question 28

31. An infinitesimally small bar magnet of dipole moment M is pointing and moving with the speed v in the positive x-direction. A small closed circular conducting loop of radius a and negligible self-inductance lies in the y-z plane with its centre 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 centre of the loop is much greater than a.

(1997C, 5M)

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Answer:

Correct Answer: 31. F=214μ02M2a4vRx8 (repulsion)

Solution:

  1. Given that x»a.

Magnetic field at the centre of the coil due to the bar magnet is

B=μ04π2Mx3=μ02πMx3

Due to this, magnetic flux linked with the coil will be,

φ=BS=μ02πMx3(πa2)=μ0Ma22x3

Induced emf in the coil, due to motion of the magnet is

e=dφdt=μ0Ma22ddt1x3=μ0Ma223x4dxdt=32μ0Ma2x4vdxdt=v

Therefore, induced current in the coil is

i=eR=32μ0Ma2Rx4v

Magnetic moment of the coil due to this induced current will be,

M=iS=32μ0Ma2Rx4v(πa2)M=32μ0πMa4vRx4

Potential energy of M in B will be

U=MBcos180U=MB=32μ0πMa4vRx4μ02πMx3

U=34μ02M2a4vR1x7F=dUdx=214μ02M2a4vRx8

Positive sign of F implies that there will be a repulsion between the magnet and the coil.



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