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A conducting current-carrying plane is p...

A conducting current-carrying plane is placed in an external uniform magnetic field. As a result, the magnetic induction becomes equal to `B_1` on one side of the plane and `B_2` on the other side. Find the magnetic force acting per unit area of the plane.

Text Solution

Verified by Experts

The correct Answer is:
`(B_1^2-B_2^2)/(2mu_0)`
Let external magnetic field is B and `B_0` is the magnetic field due
to plane.

Then `B_1=B+B_0, and B_2=B+B_0`
where `B_0=(mu_0J)/2`
Solving we get `B=(B_1+B_2)/2, B_0=(B_1-B_2)/2`
Force on plane `F=B(Jl_1)l_2`
Force per unit area:
`F/(l_1l_2)=BJ=(B_1+B_2)/2((2B_0)/(mu_0))=(B_1^2-B_2^2)/(2mu_0)`
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