Question

Two long straight parallel wires are 2 m apart, perpendicular to the plane of the paper.
The wire A carries a current of 9.6 A, directed into the plane of the paper. The wire B carries a current such that the magnetic field of induction at the point P, at a distance of 10/11 m from the wire B, is zero.
Find
(a) The magnitude and direction of the current in B.
(b) The magnitude of the magnetic field of induction at the point S.
(c) The force per unit length on the wire B.

Answer 1

(a) Direction of current at B should be
perpendicular to paper outwards. Let
current in this wire be $i_B.$ Then,
$\frac{{\mu }_0}{2\pi }\frac{i_A}{(2+\frac{10}{11})}=\frac{{\mu }_0}{2\pi }\frac{i_B}{(10/11)}$
$or\frac{i_B}{i_A}=-\frac{10}{32}$
$or{i}_B=\frac{10}{32}\times i_A=\frac{10}{32}\times 9.6=3A$
(b) Since, $A{S}^2+B{S}^2=A{B}^2$
$\therefore \angle ASB=90^{\circ }$
At S : $B_1$ = Magnetic field due to $i_A$
$\frac{{\mu }_0}{2\pi }\frac{i_A}{1.6}$
$\frac{(2\times 10^{-7})(9.6)}{1.6}$
$=12\times 10^{-7}T$
$B_2$ = Magnetic field due to $i_B$
$=\frac{{\mu }_0}{2\pi }\frac{i_B}{1.2}$
$\frac{(2\times 10^{-7})(3)}{1.2}=5\times 10^{-7}T$
Since, $B_{1}and{B}_2$ are mutually perpendicular. Net
magnetic field at 5 would be :
$B=\sqrt{B_1^2+B_2^2}$
$=\sqrt{(12\times 10^{-7}{)}^2+(5\times 10^{-7}{)}^2}$
$=13\times 10^{-7}T$
(c) Force per unit length on wire B :
$\frac{F}{l}=\frac{{\mu }_0}{2\pi }\frac{i_A{i}_B}{r}(r=AB=2m)$
$=\frac{(2\times10^{-7)(9.6\times3)}{2}$
$= 2.88 \times 10^{-6} N/m$