One long wire lies along an xaxis and carries a current of$\mathbf{30}\mathbf{}\mathbf{A}$ in the positive xdirection. A second long wire is perpendicular to the xyplane, passes through the point $\left(\mathbf{0}\mathbf{,}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{,}\mathbf{}\mathbf{0}\right)$, and carries a current of $\mathbf{40}\mathbf{}\mathbf{A}$ in the positive zdirection. What is the magnitude of the resulting magnetic field at the point$\left(\mathbf{0}\mathbf{,}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{,}\mathbf{}\mathbf{0}\right)$?

Using Biot – Savart’s law, calculate the magnitude of the magnetic field due to both wires at the given point. As both the wires produce magnetic fields perpendicular to each other, consider those values as components and calculate the magnitude.

Formula:

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

The magnitude of the magnetic field due to the first wire at the point $\left(0,0.2\text{m},0\right)$is obtained as:

$B_1=\frac{{\mu }_{0}I}{2\pi r_1}\hat{k}$

$B_1=\frac{4\pi \times {10}^{-7}\times 30}{2\pi \times 2.0}\hat{k}$

$B_1=\left(3.0\times {10}^{-6}T\right)\hat{k}$

The distance of the second wire from the given point is:

$r_2=4.0-2.0$

$r_2=2.0\text{m}$

Consider the expression for the field as:

$B_2=\frac{{\mu }_{0}I}{2\pi r_1}\hat{i}$

$B_2=\left(4.0\times {10}^{-6}T\right)\hat{i}$

Since, localid="1663141622243" $B_1$and localid="1663141629790" $B_2$are perpendicular to each other at point localid="1663141253833" $\left(0,0.2\text{m},0\right)$.

So the resultant will be

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

$B_{net}=\sqrt{{\left(3.0\times {10}^{-6}\text{T}\right)}^2+{\left(4.0\times {10}^{-6}\text{T}\right)}^2}$

$B_{net}=5.0\times {10}^{-6}\text{T}$

Thus, the magnitude of the resultant magnetic field at point $\left(0,0.2\text{m},0\right)$will be$5.0\times {10}^{-6}T$.