A 20 cm-long wire carrying a current of 6 A is immersed in a uniform magnetic field of 3 T. If the magnetic field is oriented at an angle of 30° to the wire, what is the magnitude of the force that the wire will experience?

the  magnitude of the force that the wire will experience = 1.8 N

Explanation:

The force on a current carrying wire placed in a magnetic field is :

F = Idl × B

where:

I = current flowing through the wire

dl = length of the wire

B = magnetic field

We can equally say that :

where : sin θ is the angle at which the orientation from the magnetic field  to the wire occurs = 30°

Then;

Given that:

L = 20 cm = 0.2 m

I = 6 A

B = 3 T

θ = 30°

Then:

F = 3 × 6 × 0.2 sin 30°

F = 1.8 N

Therefore, the  magnitude of the force that the wire will experience = 1.8 N

where are the vectors ?

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