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

Answer:

Magnetic field, B = 0.74 T

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 :

$F=ILB$

$B=\dfrac{F}{IL}$

$B=\dfrac{2.2}{1.5\times 2}$

B = 0.74 T

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