A 7.00-cm-long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wire and rail is 0.330 Ω . Pulling the wire at a steady speed of 4.00 m/s causes 4.30 W of power to be dissipated in the circuit.
How big is the pulling force?
What is the strength of the magnetic field?

(a) 1.075 N

(b) 4.254 T

Explanation:

(a)

From the question,

Power = Force×Velocity.

P = Fv Equation 1

Where P = Power dissipated in the circuit, F = Pulling force of the wire, v = speed of the wire.

make F the subject of the equation

F = P/v Equation 2

Given: P = 4.30 W, v = 4.0 m/s.

Substitute into equation 2

F = 4.30/4

F = 1.075 N.

(b)

Applying,

F = BILsinФ Equation 2

Where B = Strength of the magnetic field, I = current in the wire, L = Length of the wire, Ф = angle between the wire and the magnetic field.

make B the subject of the equation

B = F/ILsinФ Equation 3

But,

P = I²R Equation 4

Where R = resistance of the wire.

make I the subject of the equation

I = √(P/R) Equation 5

Given: P = 4.30 W, R = 0.330 Ω

Substitute into equation 5

I = √(4.3/0.33)

I = √13.03

I = 3.61 A.

Also given: L = 7 cm = 0.07 m, Ф = 90°

Substitute into equation 3

B = 1.075/(0.07×3.61×sin90)

B = 1.075/0.2527

B = 4.254 T.

answer and explanation;