Chapter 20: Q66P (page 862)

In Figure 20.128 on the left is a region of uniform magnetic field $B_{1}$ into the page, and adjacent on the right is a region of uniform magnetic field $B_{2}$ also into the page. The magnetic field $B_{2}$ is smaller than $B_{1} (B_{2} < B_{1})$ . You pull a rectangular loop of wire of length w, height h, and resistance R from the first region into the second region, on a frictionless surface. While you do this you apply a constant force F to the right, and you notice that the loop doesn’t accelerate but moves with a constant speed.
Calculate this constant speed v in terms of the known quantities $B_{1}$ , $B_{2}$ , w, h, R and F , and explain your calculation carefully. Also show the approximate surface-charge distribution on the loop.

Short Answer

Expert verified

$v = \frac{F R}{h^{2} {(B_{1} - B_{2})}^{2}}$

Step by step solution

Given Data

$\begin{array}{l} magnetic field in the left region = B_{1} \\ magnetic field in the right region = B_{2} \\ Force = F \\ Resistance = R \end{array}$

Concept

Magnetic field is expressed as a field where a magnetic material is charged by the force of magnetism acts.

Calculate the constant speed

Apply Faraday’s law,

$\begin{matrix} |ξ| = \frac{d ϕ}{d f} \\ = \frac{B_{1} d A_{B 1}}{d t} + \frac{B_{2} d A_{B 2}}{d t} \\ = B_{1} h v + B_{2} h v \\ |ξ| = h v (B_{1} - B_{2}) \\ I = \frac{|ξ|}{R} \\ = \frac{h v (B_{1} - B_{2})}{R} \end{matrix}$

Force on wire,

$\begin{matrix} F_{1} = B_{1} I_{1} h \\ = \frac{h v (B_{1} - B_{2}) B_{1} h}{R} \\ F_{1} = \frac{h^{2} v (B_{1} - B_{2}) B_{1}}{R} \\ F_{2} = \frac{h^{2} v (B_{1} - B_{2}) B_{2}}{R} \end{matrix}$

Total Force

$\begin{matrix} = F_{1} - F_{2} \\ F = \frac{h^{2} v (B_{1} - B_{2}) B_{1}}{R} - \frac{h^{2} v (B_{1} - B_{2}) B_{2}}{R} \\ F = \frac{h^{2} v {(B_{1} - B_{2})}^{2}}{R} \\ v = \frac{F R}{h^{2} {(B_{1} - B_{2})}^{2}} \end{matrix}$