Question

A straight wire carries a current of 10 A at an angle of 30° with respect to the direction of a uniform 0.30-T magnetic field. Find the magnitude of the magnetic force on a 0.50-m length of the wire.

Answer 1

Answer:

Explanation:

Given that,

Current in wire is

I = 10A

And the current makes an angle of 30° with respect to the magnetic field

Then, θ = 30°

And the magnetic field is

B = 0.3 T

Length of the wire is

L = 0.5m

Force on the wire F?

The force on the wire in calculated using

F = iL × B

Where

The magnitude of the cross produce of L and B is

L × B = LB•Sinθ

Then, force becomes

F = iLB•Sinθ

F = 10 × 0.5 × 0.3 × Sin30

F = 0.75 N

The force on the wire is 0.75 Newton

Answer 2

Answer:

The magnitude of the magnetic force on the wire is 0.75 N

Explanation:

Given;

current in the wire, I = 10 A

angle of inclination to magnetic field, θ = 30°

magnetic field strength, B = 0.30-T

length of the wire, L = 0.50-m

The magnitude of magnetic force acting on the wire is given as;

F = BILSinθ

where;

B is magnetic field strength

I is the current in the wire

L is length of the wire

θ is the angle of inclination

F = 0.3 x 10 x 0.5 x sin(30)

F = 0.3 x 10 x 0.5 x 0.5

F = 0.75 N

Therefore,  the magnitude of the magnetic force on the wire is 0.75 N