Question

Find the magnetic force acting per unit area of the plane in the case illustrated in above figure.

• A. $F=\frac{B_1^2-B_2^2}{2{\mu }_0}$
• B. $F=\frac{B_2^2-B_1^2}{2{\mu }_0}$
• C. $F=\frac{B_1^2-B_2^2}{{\mu }_0}$
• D. $F=\frac{B_2^2-B_1^2}{{\mu }_0}$

Answer 1

Answer: (A)

$F=\frac{B_1^2-B_2^2}{2{\mu }_0}$

Let, $B_c$ be the induction at center.
Hence,
$B_1=B+B_c$ and $B_2=B-B_c$
$\Rightarrow B=\frac{B_1+B_2}{2}$ and $B_c=\frac{B_1-B_2}{2}$
But,
$B_c=\frac{{\mu }_{0}I}{2}$
$I=\frac{2B_c}{{\mu }_0}$
Hence,
$F=BI=B\left(\frac{2B_c}{{\mu }_0}\right)$
$F=2/{\mu }_0\left(\frac{B_1+B_2}{2}\right)\left(\frac{B_1-B_2}{2}\right)$
$F=\frac{(B_1{)}^2-(B_2{)}^2}{2{\mu }_0}$