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 $ABCD$ due to currents through $AOB$ and $COD$ are perpendicular to each other. Hence, $B={\left(B_1^2+B_2^2\right)}^{1/2}$
$={\left[{\left(\frac{{\mu }_0}{4\pi }\frac{2I_1}{d}\right)}^2+{\left(\frac{{\mu }_0}{4\pi }\frac{2I_2}{d}\right)}^2\right]}^{1/2}$
$=\frac{{\mu }_0}{2\pi d}{\left(I_1^2+I_2^2\right)}^{1/2}$