Question

A wire with a current of 3.40 A is to be formed into a circular loop of one turn. If the required value of the magnetic field at the center of the loop is 20 µT, what is the required radius?

Answer 1

Answer:

0.107 m

Explanation:

The magnetic field at the center of a current-carrying loop is given by

$B=\frac{\mu_0 I}{2r}$

where

$\mu_0$ is the vacuum permeability

I is the current

r is the radius of the loop

In this problem we have

I = 3.40 A is the current in the loop

$B=20 \mu T=20\cdot 10^{-6}T$ is the magnetic field at the centre of the loop

So, solving the formula for r we find

$r=\frac{\mu_0 I}{2B}=\frac{(12.56\cdot 10^{-7} H/m)(3.40 A)}{2(20\cdot 10^{-6} T)}=0.107 m$