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

0.107 m

Step-by-step explanation:

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

`B=(\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=(\mu_0 I)/(2B)=((12.56\cdot 10^(-7) H/m)(3.40 A))/(2(20\cdot 10^(-6) T))=0.107 m`