Question

A 10.0-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 0.300 Ω . Pulling the wire at a steady speed of 4.00 m/s causes 3.90 W of power to be dissipated in the circuit.

Answer 1

Question:

What are the magnitudes of the pulling force and the magnetic field

Answer:

The magnetic field strength is 2.7 T

The pulling force
`F_p` is 0.975 N

Step-by-step explanation:

Here we have;

Length of wire, l = 10.0 cm = 0.1 m

Resistance of wire, R = 0.300 Ω

Speed of wire, v = 4.00 m/s

Power dissipated = 3.90 W

Based on the given data, we apply the relation;

`I =(Blv)/(R)`..............(1)

Where:

B = Magnetic field strength

I = Current

Since P = I²R, we have;

3.9 = I²·0.300Ω

∴ I² = 3.9/0.300 = 13

From which I = √13

Substituting the value of I in equation (1) above, we have;

`√(13)= (B * 10.0 * 4.00)/(0.300)`

Therefore;

`B = (√(13) * 0.300)/(0.1 * 4.00) = 2.70416 \ T\approx 2.7 \ T`

The magnitude of the pulling force is given by the following relation;

`F_p = F_m = IlB = l\sqrt{(P)/(R) } (√(PR) )/(lv) =(P)/(v) = (3.9 \ W)/(4.00 \ m/s) = 0.975 \ N`

The pulling force
`F_p` = 0.975 N.