Question

Two identical conducting wires AOB and COD are placed at right angles to each other. The wire AOBcarries an electric current I${}_1$ and COD carries a current I${}_2$. The magnetic field on a point lying at adistance d from O, in a direction perpendicular to the plane of the wires AOB and COD, will be givenby

• A. $\frac{{\mu }_0}{2\pi }{\left(\frac{I_1+I_2}{d}\right)}^{1/2}$
• B. $\frac{{\mu }_0}{2\pi d}{(I{{}_1}^2+I{{}_2}^2)}^{1/2}$
• C. $\frac{{\mu }_0}{2\pi d}(I_1+I_2)$
• D. $\frac{{\mu }_0}{2\pi d}(I{{}_1}^2+I{{}_2}^2)$

Answer 1

Answer: (B)

$\frac{{\mu }_0}{2\pi d}{(I{{}_1}^2+I{{}_2}^2)}^{1/2}$

As OP is perpendicular to the plane of wires means plane of paper so the magnetic field $B_1$ due to AOB is towards left and filed $B_2$ due to COD is downwards.

Hence, the required field $B=\sqrt{B_1^2+B_2^2}$

or $B=\sqrt{(\frac{{\mu }_o{I}_1}{2\pi d}{)}^2+(\frac{{\mu }_o{I}_2}{2\pi d}{)}^2}$

$=\frac{{\mu }_0}{2\pi d}\sqrt{I_1^2+I_2^2}$