A 8.00cmlong wire is pulled along a Ushaped conducting rail in a perpendicular magnetic field. The total resistance of the wire and rail is 0.330 Omega . Pulling the wire at a steady speed of 4.0 m/s causes 4.50 W of power to be dissipated in the circuit.
A) How big is the pulling force?
B) What is the strength of the magnetic field?
2 Answers

Let B = the magnetic field
Let L = the length of the wire = 8.0 cm = 0.08 m
Let R = the resistance of the wire = 0.330 Ω
Let v = the velocity of the wire = 4 m/s
Let E = the voltage across the wire
Let P = the power of the wire = 4.5 W
The power formula is:
P = E²/R
Solving for E:
E = √{PR}
The Voltage induced by a magnetic field is:
E = vBL
Use E = E:
√{PR} = vBL
Solving for B;
B = √{PR}/vL
B = √{4.5(0.33)}/(4)(.08)
B = 3.8 Wb/m²
Let F = the pulling force
Let I = the current in the wire
P = I²R
I = √{P/R}
F = IBL
F = √{P/R}BL
F = √{4.5/.33}(3.8)(.08)
F = 1.12 N

A much simpler approach:
Power=Force*velocity
Force=power/velocity
Force= 4.5(W) / 4(m/s)
Force= 1.125 (Newtons)