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.

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

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s