Q66P In Figure 20.128 on the left is ... [FREE SOLUTION]

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鈥檛 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

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

Step by step solution

Given Data

$\begin{array}{l} 尘补驳苍别迟颈肠鈥刦颈别濒诲鈥刬苍鈥则丑别鈥刲别蹿迟鈥剅别驳颈辞苍 = B_{1} \\ 尘补驳苍别迟颈肠鈥刦颈别濒诲鈥刬苍鈥则丑别鈥剅颈驳丑迟鈥剅别驳颈辞苍 = B_{2} \\ Force = F \\ Resistance = R \end{array}$

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鈥檚 law,

$\begin{matrix} |尉| = \frac{d 蠒}{d f} \\ = \frac{B_{1} d A_{B 1}}{d t} + \frac{B_{2} d A_{B 2}}{d t} \\ = B_{1} h v + B_{2} h v \\ |尉| = h v (B_{1} 鈭� B_{2}) \\ I = \frac{|尉|}{R} \\ = \frac{h v (B_{1} 鈭� B_{2})}{R} \end{matrix}$

Force on wire,

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

Total Force

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