A bar magnet having dipole moment m is moved towards a circular loop of a conducting material having radius
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Lenz's Law
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Problem (IIT JEE 1997): An infinitesimally small bar magnet of dipole moment M⃗ 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.
Solution: Lenz's Law The magnetic field due to an infinitesimal bar magnet for end-on position is given by,
B⃗ =μ04π2Mx3ı^,
where M is magnetic moment and x is the distance from the magnet. Since distance x is much greater than radius~a of the loop, magnetic field can be taken as uniform throughout the loop. For the loop, flux ϕ, induced emf e, induced current i and the magnetic moment M′ are given by,
ϕ=B⃗ ⋅S⃗ =μ04π2Mx3(πa2)=μ0a2M2x3,e=−dϕdt=3μ0a2M2x4dxdt=3μ0a2Mv2x4,i=eR=3μ0a2Mv2Rx4,M′=iS=3πμ0a4Mv2Rx4.
According to Lenz's law, the direction of M′ is such that it opposes approaching magnet (see figure). The potential energy of the loop having magnetic moment M′ and placed in magnetic field B⃗ is,
U=−M′→⋅B⃗ =M′B=3πμ0a4Mv2Rx4μ04π2Mx3=34μ20a4M2vRx7,
and the force acting on the loop is,
F=−dUdx=214μ20a4M2vRx8.
The positive sign confirms the repulsive nature of force.
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