Question

3. I A 10-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$0.20\Omega$ . Pulling the wire at a steady speed of$4.0\text{m}/\text{s}$ causes $4.0\text{W}$of power to be dissipated in the circuit.

a. How big is the pulling force?

b. What is the strength of the magnetic field?

Short answer

Part (a) The magnitude of the pulling force is $1\text{N}$

Part (b) Strength of the magnetic field is$2.23\text{T}$

Step-by-step solution

Step 1. Given infomation

Length of the wire, $l=10\text{cm}=0.1\text{m}$

Total resistance of the wire,$R=0.20\Omega$

Steady speed,$v=4\text{m}/\text{s}$

Power dissipated,$P=4\text{W}$

Part (a)

The relation between the pulling force and the power dissipated is given as$P_{\text{dissipated}}=F_{\text{pulling}}\times v$

role="math" $\begin{array}{l}4=F_{\text{pulling}}\times 4\\ F_{\text{pulling}}=1\text{N}\end{array}$

The magnitude of the pulling force is$1\text{N}$

Part (b)

The pulling force is given by the formula

$\begin{array}{l}{F}_{\text{pull}}=\frac{v{l}^2{B}^2}{R}\\ F_{\text{pull}}=\frac{\nu l^2{B}^2}{R}\\ 1=\frac{4\times {0.1}^2\times B^2}{0.2}\\ B=2.23\text{T}\end{array}$

Strength of the magnetic field is$2.23\text{T}$