Answer:
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 :
[tex]|F| = IdlBsin \theta[/tex]
where : sin θ is the angle at which the orientation from the magnetic field to the wire occurs = 30°
Then;
[tex]|F| = B\ I \ L \ sin \theta[/tex]
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