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

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;

$I =\frac{Blv}{R}$..............(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;

$\sqrt{13}= \frac{B \times 10.0 \times 4.00}{0.300}$

Therefore;

$B = \frac{\sqrt{13} \times 0.300}{0.1 \times 4.00} = 2.70416 \ T\approx 2.7 \ T$

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

$F_p = F_m = IlB = l\sqrt{\frac{P}{R} } \frac{\sqrt{PR} }{lv} =\frac{P}{v} = \frac{3.9 \ W}{4.00 \ m/s} = 0.975 \ N$

The pulling force $F_p$ = 0.975 N.