Question

Two identical long conducting wires $AOB$ and $COD$ are placed at right angle to each other, with one above other such that ${}^{\prime }{O}^{\prime }$ their common point for the two. The wires carry $I_1\text{and}{I}_2$ currents respectively. Point ${}^{\prime }{P}^{\prime }$ is lying at distance ${}^{\prime }{d}^{\prime }$ from ${}^{\prime }{O}^{\prime }$ along a direction perpendicular to the plane containing the wires. The magnetic field at the point ${}^{\prime }{P}^{\prime }$ will be

• A. $\frac{{\mu }_0}{2\pi d}\left(\frac{I_1}{I_2}\right)$
• B. $\frac{{\mu }_0}{2\pi d}(I_1+I_2)$
• C. $\frac{{\mu }_0}{2\pi d}(I_1^2-I_2^2)$
• D. $\frac{{\mu }_0}{2\pi d}(I_1^2+I_2^2{)}^{\frac{1}{2}}$

Answer 1

The correct option is D $\frac{{\mu }_0}{2\pi d}(I_1^2+I_2^2{)}^{\frac{1}{2}}$


Magnetic field due to wire carrying current $i_1$ at $P$ is $\overrightarrow{B_1}=\frac{{\mu }_0{I}_1}{2\pi d}\left(\hat{i}\right)$
Magnetic field due to wire carrying current $i_2$ at $P$ is $\overrightarrow{B_2}=\frac{{\mu }_0{I}_2}{2\pi d}(-\hat{j})$

Resultant magnetic field at point P is
$|\overrightarrow{B}|=\sqrt{(|\overrightarrow{B_1}{|}^2+|\overrightarrow{B_2}{|}^2)}$
$|\overrightarrow{B}|={({\left(\frac{{\mu }_0{I}_1}{2\pi d}\right)}^2+{\left(\frac{{\mu }_0{I}_2}{2\pi d}\right)}^2)}^{\frac{1}{2}}$
$|\overrightarrow{B}|=\frac{{\mu }_0}{2\pi d}\sqrt{I_1^2+I_2^2}$