Question

A current carrying wire of length 37 cm is positioned perpendicular to a uniform magnetic field. If the current is 3.2 A and it is determined that there is a resultant force of 2.5 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 2.11 T.

Step-by-step explanation:

It is given that,

Length of the wire, L = 37 cm = 0.37 m

Current, I = 3.2 A

Force, F = 2.5 N

Magnetic force is given by :

`F=qvB\ sin\theta`

Here,
`\theta=90,sin\ \theta=1`

`F=qvB`

We know that,

Velocity,
`v=(L)/(t)`

`F=(qLB)/(t)`

Since,
`(q)/(t)=I`

`F=ILB`

`B=(F)/(IL)`

`B=(2.5\ N)/(3.2\ A* 0.37\ m)`

B = 2.11 T

So, the magnetic field strength is 2.11 T. Hence, this is the required solution.