Question

A wire is carrying 3.0 A of current. 4.2 cm of the wire crosses perpendicularly between the two poles of a strong horseshoe magnet. When the current is turned on, the wire experiences 0.47 N of force. What is the strength of the magnetic field?

Answer 1

The strength of the magnetic field is 3.73 T.

Step-by-step explanation:

We have,

Current in the wire is 3 A

Length of wire, L = 4.2 cm = 0.042 m

A wire crosses perpendicularly between the two poles of a strong horseshoe magnet.

When the current is turned on, the wire experiences 0.47 N of force, F = 0.47 N

It is required to find the strength of the magnetic field. The magnetic force acting on the wire is given by :

`F=iLB\sin\theta`

B is strength of the magnetic field

`B=(F)/(iL)\\\\B=(0.47)/(3* 0.042)\\\\B=3.73\ T`

So, the strength of the magnetic field is 3.73 T.