Question

Two long, parallel wires separated by 0.328 m carry equal currents of 9.812 A in the same direction. Find the magnitude of the net magnetic field 0.832 m away from each wire on the side opposite the other wire . Express your answer in microTesla.

Answer 1

Given:

The distance between the wires is,

$d=0.328\text{ m}$

The currents in each wire are in the same direction and the magnitude is,

$i=9.812\text{ A}$

To find:

The net magnetic field 0.832 m away from each wire on the side opposite the other wire

Explanation:

The magnetic field due to a current-carrying wire is,

$\begin{gathered} B=\frac{\mu_0i}{2\pi r} \\ \mu_0=4\pi\times10^{-7}\text{ H/m} \end{gathered}$

If we see the diagram of both the wires,

the point x is at a distance (x+d) from the second wire

The magnetic field due to wire 1 and wire 2 will be out of the page.

The net magnetic field at point x is,

$\begin{gathered} B=\frac{\mu_0i}{2\pi x}+\frac{\mu_0\imaginaryI}{2\pi(x+d)} \\ =\frac{4\pi\times10^{-7}\times9.812}{2\pi\times0.832}+\frac{4\pi\times10^{-7}\times9.812}{2\pi\times(0.832+0.328)} \\ =2.36\times10^{-6}+1.69\times10^{-6} \\ =4.05\times10^{-6}\text{ T} \\ =4.05\text{ }\mu T \end{gathered}$

Similarly, on the right of wire 2, the magnetic field will be the same but into the page.

Hence, the required magnetic field is,

$4.05\text{ }\mu T$