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 …

Question

asked Mar 26, 2020 107k views

1 Answer

Corasan · Answer 1 · 2020-04-02T01:24:10+0000

Answer:

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

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
. Hence, this is the required solution.