Question

A straight segment of wire has a length of 25 cm and carries a current of 5A. If the wire is perpendicular to the magnetic field of 0.60Tesla, then what is the magnitude of the magnetic force?

Answer 1

Answer:

The magnitude of the magnetic force acting on the conductor is 0.75 Newton

Explanation:

The parameters given in the question are;

The length of the straight segment of wire, L = 25 cm = 0.25 m

The current carried in the wire, I = 5 A

The orientation of the wire with the magnetic field = Perpendicular

The strength of the magnetic field in which the wire is located, B = 0.60 T

The magnetic force, 'F', is given by the following formula;

F = $\underset{I}{\rightarrow }$·L×$\underset{B}{\rightarrow }$ = I·L·B·sin(θ)

Where;

$\underset{I}{\rightarrow }$ = The current flowing, I

L = The length of the wire

$\underset{B}{\rightarrow }$ = The magnetic field strength, B

θ = The angle of inclination of the conductor to the magnetic field

Where I = 5 A, L = 0.25 m, B = 0.60 T, and θ = 90°, we get;

F = 5 A × 0.25 m × 0.60 T × sin(90°) = 0.75 N

Therefore

The magnitude of the magnetic force, F = 0.75 N.