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

Step-by-step explanation:

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

`\begin{gathered} B=(\mu_0i)/(2\pi r) \\ \mu_0=4\pi*10^(-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=(\mu_0i)/(2\pi x)+(\mu_0\imaginaryI)/(2\pi(x+d)) \\ =(4\pi*10^(-7)*9.812)/(2\pi*0.832)+(4\pi*10^(-7)*9.812)/(2\pi*(0.832+0.328)) \\ =2.36*10^(-6)+1.69*10^(-6) \\ =4.05*10^(-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`