Question

A straight wire that is 0.60 m long is carrying a current of 2.0 A. It is placed in a uniform magnetic field of strength 0.30 T. If the wire experiences a force of 0.18 N, what angle does the wire make with respect to the magnetic field?

Answer 1

The angle the wire make with respect to the magnetic field is 30°

Step-by-step explanation:

It is given that,

Length of straight wire, L = 0.6 m

Current carrying by the wire, I = 2 A

Magnetic field, B = 0.3 T

Force experienced by the wire, F = 0.18 N

Let θ be the angle the wire make with respect to the magnetic field. Magnetic force is given by :

`F=ILB\ sin\theta`

`\theta=sin^(-1)((F)/(ILB))`

`\theta=sin^(-1)((0.18\ N)/(2\ A* 0.6* 0.3\ T))`

`\theta=30^(\circ)`

So, the angle the wire make with respect to the magnetic field is 30°. Hence, this is the required solution.