The diameter of a turns circular coil 8.0cm and it has a current of 3.0A (I) find magnetic moment of coil?

To find the magnetic moment of the coil, we can use the formula:

Magnetic moment (m) = current (I) * area (A) * number of turns (N)

First, we need to find the area of the coil. The area of a circle is given by the formula:

A = πr^2

Given that the diameter of the coil is 8.0 cm, the radius (r) is half of the diameter, so r = 4.0 cm = 0.04 m.

Therefore, A = π(0.04)^2 = 0.005 m^2

Now, we can calculate the magnetic moment:

m = 3.0 A * 0.005 m^2 * 1 (since there is only one turn in the coil)

m = 0.015 A m^2

Therefore, the magnetic moment of the coil is 0.015 A m^2.

The magnetic moment of a current loop is given by:

```
μ = I - A
```

where:

- μ is the magnetic moment (in A⋅m²)
- I is the current (in A)
- A is the area of the loop (in m²)

In this case, the diameter of the coil is 8.0 cm, so the radius is 4.0 cm or 0.04 m. The area of the coil is then:

```
A = πr² = π(0.04 m)² = 0.0050 m²
```

Substituting these values into the equation for magnetic moment, we get:

```
μ = I - A = 3.0 A - 0.0050 m² = 0.015 A⋅m²
```

Therefore, the magnetic moment of the coil is 0.015 A⋅m².