Question

A conductor of length 2m carrying a 1.5A current experiences a force of 2.2 N. What is the magnitude of the magnetic field that acts perpendicular to the flow of current in the conductor?

Answer 1

Magnetic field, B = 0.74 T

Step-by-step explanation:

It is given that,

Length of the conductor, L = 2 m

Current in conductor, I = 1.5 A

Magnetic force, F = 2.2 N

We need to find the magnitude of the magnetic field that acts perpendicular to the flow of current in the conductor. It is given by :

`F=ILB`

`B=(F)/(IL)`

`B=(2.2)/(1.5* 2)`

B = 0.74 T

So, the magnitude of magnetic field will be 0.74 T. Hence, this is the required solution.