A circular coil consists of N = 410 closely winded turns of wire and has a radius R = 0.75 m. A counterclockwise current I = 2.4 A is in the coil. The coil is set in a magnetic field of magnitude B = 1.1 T. The magnetic field points to right

The complete question is;

A circular coil consists of N = 410 closely winded turns of wire and has a radius R = 0.75 m. A counterclockwise current I = 2.4 A is in the coil. The coil is set in a magnetic field of magnitude B = 1.1 T.

a. Express the magnetic dipole moment μ in terms of the number of the turns N, the current I, and radius

R.

b. Which direction does μ go?

A) μ = 1738.87 A.m²

B) The direction of the magnetic moment will be in upward direction.

Explanation:

We are given;

The number of circular coils;

N = 410

The radius of the coil;R = 0.75m

The current in the coils; I = 2.4 A

The strength of magnetic field;

B =1.1T

The formula for magnetic dipole moment is given as;

μ = NIA

Where;

N is number of turns

I is current

A is area

Now, area; A = πr²

So, A = π(0.75)²

Thus,plugging in relevant values, the magnetic dipole moment is;

μ = 410 * 2.4 * π(0.75)²

μ = 1738.87 A.m²

B) According to Fleming's right hand rule, the direction of the magnetic moment comes out to be in upward direction.