Question

Two identical long conducting wires $AOB$ and $COD$ are placed at right angle to each other, with one above other such that $O$ is their common point for the two. The wires carry $I_1$ and $I_2$ currents, respectively. Point $P$ is lying at distance $d$ from $O$ along a direction perpendicular to the plane containing the wires. The magnetic field at the point $P$ will be

• A. $\frac{\mu_{0}}{2 \pi d}\left(\frac{I_{1}}{I_{2}}\right)$
• B. $\frac{\mu_{0}}{2 \pi d}\left(I_{1}+I_{2}\right)$
• C. $\frac{\mu_{0}}{2 \pi d}\left(I_{1}^{2}-I_{2}^{2}\right)$
• D. $\frac{\mu_{0}}{2 \pi d}\left(I_{1}^{2}+I_{2}^{2}\right)^{I / 2}$

Answer 1

Answer: (D)

The magnetic field at the point $P$, at a perpendicular distance $d$ from $O$ in a direction perpendicular to the plane $A B C D$ due to currents through $A O B$ and $C O D$ are perpendicular to each other. Hence, $B=\left(B_{1}^{2}+B_{2}^{2}\right)^{1 / 2}$
$=\left[\left(\frac{\mu_{0}}{4 \pi} \frac{2 I_{1}}{d}\right)^{2}+\left(\frac{\mu_{0}}{4 \pi} \frac{2 I_{2}}{d}\right)^{2}\right]^{1 / 2}$
$=\frac{\mu_{0}}{2 \pi d}\left(I_{1}^{2}+I_{2}^{2}\right)^{1 / 2}$