Question

D) A wire carrying a current of 10A and 2m in length is placed in a field of flux density

Answer 1

1) The force on the wire is 1.5 N

2) The force on the wire is zero

Explanation:

1)

The force experienced by a current-carrying wire in a magnetic field is given by

$F=ILB sin \theta$

where

I is the current in the wire

L is the length of the wire

B is the magnetic flux density

$\theta$ is the angle between the direction of the wire and of the field

In this part of the problem we have

I = 10 A

L = 2 m

B = 0.15 T

$\theta=30^{\circ}$

Substituting, we find

$F=(10)(2)(0.15)(sin 30^{\circ})=1.5 N$

2)

As before, the force experienced by the wire is

$F=ILB sin \theta$

where

I is the current in the wire

L is the length of the wire

B is the magnetic flux density

$\theta$ is the angle between the direction of the wire and of the field

In this case, however, the angle between the wire and the field is

$\theta=0^{\circ}$

And therefore

$sin 0^{\circ} = 0$

which means that the wire does not experience any force. In fact, the force experienced by the wire is zero when the wire is parallel to the field (such as in this case) and maximum when the wire is perpendicular to the field.

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