Question

Two identical conducting wires $AOB$ and $COD$ are placed at right angles to each other. The wire $AOB$ carries an electric current $I_1$ and $COD$ carries a current $I_2$ . The magnetic field on a point lying at a distance $d$ from $O$ , in a direction perpendicular to the plane of the wires $AOB$ and $COD,$ will be given by

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

Answer 1

Answer: (B)

The magnetic field induction at a point $P_1$ at a distance $d$ from $O$ in a direction perpendicular to the plane of the wires and due to the currents through $AOB$ and $COD$ as shown, are perpendicular to each other and it is
Hence, $B=\sqrt{B_1^2+B_2^2}$
$\Rightarrow B={\left(\left[{\left(\frac{{\left(\mu \right)}_0}{4\pi }\frac{2I_1}{d}\right)}^2+{\left(\frac{{\left(\mu \right)}_0}{4\pi }\frac{2I_2}{d}\right)}^2\right]\right)}^{\frac{1}{2}}$
$\therefore B=\frac{{\left(\mu \right)}_0}{2\pi d}\sqrt{\left(I_1^2+I_2^2\right)}$