IIT JEE , AIEEE Forum - Physics , Magnetism

bumba

a circular loop of radius a carrying current i is placed in a surface where magnetic field is perpendicular to the plane and going through it.calculate the force on a small part dl and the force of compression in the wire?????????????????

devesh_l2k007

Nice prob DUde !!
But has got an easy soln
dF=idl B
Net force on the whole loop wud be zero
but body would stretch (apply the cross prod at every pt )
Hence the force of compression wud be -idl B

bumba

u r answer to the first part is correct but to the 2nd part its wrong.the answer given is iab.

iitkgp_bipin

Force on small element = i(

dl)xB = i(dl)B

For a closed loop (

dl) = 0
Hence net force is 0.

Now let the small element subtend an angle d

at the centre.

Let the tension be T in the wire.
Draw free body diagram of the element.

The tangential forces Tcos(d

/2) cancel out.

The radial forces act towards centre and add up which is balanced by the magnetic force.
2Tsin(d

/2) = i(dl)B

Since d

is small sin(d

/2) can be approximated as d

/2.
2T(d

/2) = i(dl)B

From geometry, arc length dl = ad

(where a is the radius)

2T(d

/2) = i(ad

)B
T = iaB

umang

I agree with ur solution sir . But here's a shorter technique .

Let us take the semicircular part of the circle . Then , the two tension forces are downwards (if u take the upper half) and magnetic force is upwards . Then , balance the forces and u get the answer .

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