Question

Two long identical conducting wires $\text{AOB}$ and $\text{COD}$ are placed at right angles to each other. The wire $\text{AOB}$ carries an electric current $I_1$ and $\text{COD}$ carries a current $I_2$. Find the magnetic field at point $\text{P}$ at a distance $d$ from the intersection point $\text{O}$, in a direction perpendicular to the plane of the wires $\text{AOB}$ and $\text{COD}$ is-

• A. $\frac{{\mu }_0}{2\pi }\sqrt{\frac{I_1+I_2}{d}}$
• B. $\frac{{\mu }_0}{2\pi d}\sqrt{I_1^2+I_2^2}$
• 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

The correct option is B $\frac{{\mu }_0}{2\pi d}\sqrt{I_1^2+I_2^2}$

Let us consider the wire $\text{AOB}$ and $\text{COD}$ as (1) and (2) and their field at point $\text{P}$ be $B_1$ and $B_2$ respectively.


From the above diagram, we can see that magnetic field $B_1$ and $B_2$ are perpendicular to each other.

Field due to long current carrying conductor is,

$B=\frac{{\mu }_{0}I}{2\pi d}$

$\Rightarrow B_1=\frac{{\mu }_0{I}_1}{2\pi d}$

$\Rightarrow B_2=\frac{{\mu }_0{I}_2}{2\pi d}$

$B_{net}=\sqrt{B_1^2+B_2^2}=\frac{{\mu }_0}{2\pi d}\sqrt{I_1^2+I_2^2}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{B}\right)$ is the correct answer.