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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.
1071567_613df880bcfc46db8f6a658d0964ffaf.png

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Solution

(a) The external field must be B1+B22, which when superposed with the internal field B1B22 (of opposite sign on the two sides of the plate) must given actual field. Now
B1B22=12μ0i
or, i=B1B2μ0
Thus, F=B21B222μ0
(b) Here, the external field must be B1B22 upward with an internal field, B1+B22, upward on the left and downward on the right. Thus,
i=B1+B2μ0 and F=B21B222μ0
(c) Out boundary condition following from
Gauss's law is, B1cosθ1=B2cosθ2.
Also, (B1sinθ1+B2sinθ2)=μ0i where
i= current per unit length.
The external field parallel to the plate must be B1sinθ1B2sinθ22
(The perpendicular component B1cosθ1, does not matter since the corresponding force is tangential)
Thus, F=B21sin2θ1B22sin2θ22μ0 per unit area
=B21B222μ0 per unit area.
The direction of the current in the plane conductor is perpendicular to the paper and beyond the drawing.
1797829_1071567_ans_32123e5c3272429ebafe7bdd74c39e00.png

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