Question

A magnetic field of magnitude 0.550 T is directed parallel to the plane of a circular loop of radius 43.0 cm. A current of 5.80 mA is maintained in the loop. What is the magnetic moment of the loop? (Enter the magnitude.)

Answer 1

`\mu = 3.36* 10^(-3)\ A-m^2`

Step-by-step explanation:

Given that,

The magnitude of magnetic field, B = 0.55 T

The radus of the loop, r = 43 cm = 0.43 m

The current in the loop, I = 5.8 mA = 0.0058 A

We need to find the magnetic moment of the loop. It is given by the relation as follows :

`\mu = AI\\\\\mu=\pi r^2* I`

Put all the values,

`\mu=\pi * (0.43)^2* 0.0058\\\\=3.36* 10^(-3)\ A-m^2`

So, the magnetic moment of the loop is equal to
`3.36* 10^(-3)\ A-m^2`.