Question

A loop of wire is carrying current of 2 A . The radius of the loop is 0.4 m. What is the magnetic field at a distance 0.09 m along the axis and above the center of the loop

Answer 1

`B=2.91\ \mu T`

Step-by-step explanation:

Given that,

The current in the loop, I = 2 A

The radius of the loop, r = 0.4 m

We need to find the magnetic field at a distance 0.09 m along the axis and above the center of the loop. The formula for the magnetic field at some distance is given as follows :

`B=(\mu_o)/(4\pi )(2\pi r^2 I)/((r^2+d^2)^(3/2))`

Put all the values,

`B=10^(-7)* (2\pi * 0.4^2 * 2)/((0.4^2+0.09^2)^(3/2))\\\\=2.91* 10^(-6)\ T\\\\or\\\\B=2.91\ \mu T`

So, the required magnetic field is equal to
`2.91\ \mu T`.