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

Answer:

$B=2.91\ \mu T$

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 :

$B=\dfrac{\mu_o}{4\pi }\dfrac{2\pi r^2 I}{(r^2+d^2)^{3/2}}$

Put all the values,

$B=10^{-7}\times \dfrac{2\pi \times 0.4^2 \times 2}{(0.4^2+0.09^2)^{3/2}}\\\\=2.91\times 10^{-6}\ T\\\\or\\\\B=2.91\ \mu T$

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