Question

An electromagnet produces a magnetic field of 0.530 T in a cylindrical region of radius 2.80 cm between its poles. A straight wire carrying a current of 10.7 A passes through the center of this region and is perpendicular to both the axis of the cylindrical region and the magnetic field. What magnitude of force does this field exert on the wire?

Answer 1

Magnetic force, F = 0.317 N

Step-by-step explanation:

Given that,

Magnetic field produced by an electromagnet, B = 0.53 T

Radius of the electromagnet, r = 2.8 cm = 0.028 m

Current, I = 10.7 A

Let F is the magnitude of force this field exert on the wire. Its formula is given by :

`F=ILB\ sin\theta`

Here,
`\theta=90^(\circ)`

`F=ILB\ sin(90)`

`F=ILB` (L = 2r)

`F=10.7* 2* 0.028* 0.53`

F = 0.317 N

So, the magnitude of force the field exert on the wire is 0.317 N. Hence, this is the required solution.