Question

Two infinite length wire carries currents $8\text{A}$ and $6\text{A}$ respectively and placed along $x$ and $y$- axis. Magnetic field at a point $\text{P}(0,0,d)\text{m}$ will be:

• A. $\frac{7{\mu }_0}{\pi d}$
• B. $\frac{10{\mu }_0}{\pi d}$
• C. $\frac{14{\mu }_0}{\pi d}$
• D. $\frac{5{\mu }_0}{\pi d}$

Answer 1

The correct option is D $\frac{5{\mu }_0}{\pi d}$

The direction of magnetic field due to infinite length of wires at point $\text{P}$ is shown in figure which is obtained from right- hand thumb rule.

Magnetic field due to wire $1$ is

$B_1=\frac{{\mu }_0{i}_1}{2\pi d_1}(-\hat{j})$

$B_1=\frac{{\mu }_0\times 8}{2\pi \times d}(-\hat{j})=-\frac{4{\mu }_0}{\pi d}\hat{j}$

Similarly, due to wire $2$,

$B_2=\frac{{\mu }_0{i}_2}{2\pi d}\left(\hat{i}\right)$

$B_2=\frac{{\mu }_0\times 6}{2\pi d}\hat{i}=\frac{3{\mu }_0}{\pi d}\hat{i}$

So the net magnetic field at point $\text{P}$ is

$B_{net}=\sqrt{|B_1{|}^2+|B_2{|}^2}$

$B_{net}=\sqrt{{\left(\frac{-4{\mu }_0}{\pi d}\right)}^2+{\left(\frac{3{\mu }_0}{\pi d}\right)}^2}$

$\therefore B_{net}=\frac{5{\mu }_0}{\pi d}$

Therefore, the correct option is $\left(d\right)$.