Question

A current-carrying wire of length 52.0 cm is positioned perpendicular to a uniform magnetic field. If the current is 15.0 A and it is determined that there is a resultant force of 2.3 N on the wire due to the interaction of the current and field, what is the magnetic field strength?

Answer 1

The magnetic field strength is 0.29 T.

Step-by-step explanation:

Given that,

Length of current-carrying wire, L = 52 cm = 0.52 m

Current, I = 15 A

Magnetic force, F = 2.3 N

We need to find the magnetic field strength. We know that the magnetic force is given by :

`F=ILB`

B is magnetic field strength

`B=(F)/(IL)\\\\B=(2.3)/(0.52* 15)\\\\B=0.29\ T`

So, the magnetic field strength is 0.29 T.