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

Answer:

$\mu = 3.36\times 10^{-3}\ A-m^2$

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 :

$\mu = AI\\\\\mu=\pi r^2\times I$

Put all the values,

$\mu=\pi \times (0.43)^2\times 0.0058\\\\=3.36\times 10^{-3}\ A-m^2$

So, the magnetic moment of the loop is equal to$3.36\times 10^{-3}\ A-m^2$.