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?
Answer:
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.
