Question

The area vector of a square loop of 5 turns of a conductor each with side length of 0.2 m carrying a current of 2 A is antiparallel to a uniform magnetic field of 50.0 T in which it lies.

Answer 1

the magnitude and the direction of the total magnetic field is 0.4 Am² antiparallel to the area vector

Step-by-step explanation:

Given that:

The area vector of a square loop has 5 numbers of turns i.e n = 5

each with side length = 0.2 m

Current I = 2 A

uniform magnetic field = 50.0 T

Now; the magnitude of the total magnetic field B is calculated as :

B = IA

where;

I = current

A = area ( n × l²)

B = I ( n × l²)

B = 2 × 5 × 0.2²

B = 0.4 Am²

The direction of the magnetic moment is antiparallel to the area vector;

Hence ; the magnitude and the direction of the total magnetic field is 0.4 Am² antiparallel to the area vector