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?

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 = ·L× = I·L·B·sin(θ)

Where;

= The current flowing, I

L = The length of the wire

= 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.

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.