1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# A conducting current-carrying plane is placed in an external uniform magnetic field. As a result, the magnetic induction becomes equal to B1 on the one side of the planes and to B2 on the other. Find the magnetic force acting per unit area of the plane in the cases illustrated in Fig. Determine the direction of the current in the plane in each case.

Open in App
Solution

## (a) The external field must be B1+B22, which when superposed with the internal field B1−B22 (of opposite sign on the two sides of the plate) must given actual field. NowB1−B22=12μ0ior, i=B1−B2μ0Thus, F=B21−B222μ0(b) Here, the external field must be B1−B22 upward with an internal field, →B1+B22, upward on the left and downward on the right. Thus,i=B1+B2μ0 and F=B21−B222μ0(c) Out boundary condition following fromGauss's law is, B1cosθ1=B2cosθ2.Also, (B1sinθ1+B2sinθ2)=μ0i wherei= current per unit length.The external field parallel to the plate must be B1sinθ1−B2sinθ22(The perpendicular component B1cosθ1, does not matter since the corresponding force is tangential)Thus, F=B21sin2θ1−B22sin2θ22μ0 per unit area=B21−B222μ0 per unit area.The direction of the current in the plane conductor is perpendicular to the paper and beyond the drawing.

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Self and Mutual Inductance
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program