Question

A loop of wire with cross-sectional area 1×10^−3 m^2 lays centered in the xy -plane. The wire carries a uniform current of 180A running counter-clockwise. What is the magnitude of the magnetic moment of the current loop?

Answer 1

`\mu=180* 10^(-3)A-m^2`

Step-by-step explanation:

Given that,

Area of the loop,
`A=10^(-3)\ m^2`

Current flowing in the wire, I = 180 A

We need to find the magnetic moment of the current loop. It is given by :

`\mu=I* A`

`\mu=180* 10^(-3)`

`\mu=180* 10^(-3)A-m^2`

So, the magnetic moment of the current loop is
`180* 10^(-3)A-m^2`. Hence, this is the required solution.