Question

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 the one side of the planes and to $B_2$ 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.

Answer 1

(a) The external field must be $\frac{B_1+B_2}{2}$, which when superposed with the internal field $\frac{B_1-B_2}{2}$ (of opposite sign on the two sides of the plate) must given actual field. Now
$\frac{B_1-B_2}{2}=\frac{1}{2}{\mu }_{0}i$
or, $i=\frac{B_1-B_2}{{\mu }_0}$
Thus, $F=\frac{B_1^2-B_2^2}{2{\mu }_0}$
(b) Here, the external field must be $\frac{B_1-B_2}{2}$ upward with an internal field, $\overrightarrow{B_1+B_2}2$, upward on the left and downward on the right. Thus,
$i=\frac{B_1+B_2}{{\mu }_0}$ and $F=\frac{B_1^2-B_2^2}{2{\mu }_0}$
(c) Out boundary condition following from
Gauss's law is, $B_1\cos {\theta }_1=B_2\cos {\theta }_2$.
Also, $(B_1\sin {\theta }_1+B_2\sin {\theta }_2)={\mu }_{0}i$ where
$i=$ current per unit length.
The external field parallel to the plate must be $\frac{B_1\sin {\theta }_1-B_2\sin {\theta }_2}{2}$
(The perpendicular component $B_1\cos {\theta }_1$, does not matter since the corresponding force is tangential)
Thus, $F=\frac{B_1^2{\sin }^2{\theta }_1-B_2^2{\sin }^2{\theta }_2}{2{\mu }_0}$ per unit area
$=\frac{B_1^2-B_2^2}{2{\mu }_0}$ per unit area.
The direction of the current in the plane conductor is perpendicular to the paper and beyond the drawing.