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Chapter 30: Q. 46 (page 872)

FIGURE shows a 4.0-cm-diameter loop with resistance 0.10鈩�around a 2.0-cm-diameter solenoid. The solenoid is 10 cm long, has 100turns, and carries the current shown in the graph. A positive current is cwwhen seen from the left. Find the current in the loop at(a)t=0.5s,(b)t=1.5s,and(c)t=2.5s.

Short Answer

Expert verified

(a) By assign,t=0.5鈫�Iinduced=0A

(b) By assign,t=1.5鈫�Iinduced=12.56脳10-5A

(c) By assign,t=2.5鈫�Iinduced=0A

Step by step solution

01

Introduction

Given values:

R=0.1惟

n=1000

r=R'=2脳10-2m

r=1脳10-2

Bs=渭onI

Use next equations,

E(2蟺R)=蟺r2dBdt

E2蟺R'=蟺r2渭ondIdt

E=蟺r22蟺R'渭ondIdt

E=r2渭on2R'dIdt

Iinduced=10-222脳2脳10-2脳4蟺脳10-7脳1000dIdt=3.14脳10-6dIdt

02

Find Iinduced

a.) Iinducedat t=0.5:

Iinduced=3.14脳10-6脳0

Iinduced=0A鈫�dIdt=0

b.) Iinducedat t=1.5:

Iinduced=3.14脳10-6脳40

Iinduced=12.56脳10-5A鈫�dIdt=40

c.) Iinducedat t=2.5:

Iinduced=0A鈫�dIdt=0

03

Explain with graph

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Most popular questions from this chapter

Let's look at the details of eddy-current braking. A square CALC loop, lengthlon each side, is shot with velocityv0into a uniform magnetic field localid="1648921142252" B. The field is perpendicular to the plane of the loop. The loop has mass localid="1648921150406" mand resistancelocalid="1648921154874" R,and it enters the field atlocalid="1648921174281" t=0s. Assume that the loop is moving to the right along thelocalid="1648921181007" x-axis and that the field begins atlocalid="1648921198444" x=0m.

a. Find an expression for the loop's velocity as a function of time as it enters the magnetic field. You can ignore gravity, and you can assume that the back edge of the loop has not entered the field.

b. Calculate and draw a graph oflocalid="1648921211473" vover the intervallocalid="1648921223129" 0s鈮�t鈮�0.04sfor the case thatlocalid="1648816410574" width="87">v0=10m/s,localid="1648921234041" l=10cm,localid="1648921244487" m=1.0g,localid="1648921254639" R=0.0010惟,and localid="1648921264943" B=0.10T. The back edge of the loop does not reach the field during this time interval.

I

I The metal equilateral triangle in FIGURE EX30.11,20cmon each side, is halfway into a0.10Tmagnetic field.

a. What is the magnetic flux through the triangle?

b. If the magnetic field strength decreases, what is the direction of the induced current in the triangle?

FIGURE shows a U-shaped conducting rail that is oriented vertically in a horizontal magnetic field. The rail has no electric resistance and does not move. A slide wire with mass mand resistance Rcan slide up and down without friction while maintaining electrical contact with the rail. The slide wire is released from rest.

a. Show that the slide wire reaches a terminal speed vterm, and find an expression forvterm .

b. Determine the value of vtermif l=20cm,m=10g,R=0.10惟and B=0.50T

Your camping buddy has an idea for a light to go inside your tent. He happens to have a powerful and heavy horseshoe magnet that he bought at a surplus store. This magnet creates a 0.20Tfield between two pole tips 10cmapart. His idea is to build the hand-cranked generator shown in FIGURE .He thinks you can make enough current to fully light a 1.0鈩�lightbulb rated at 4.0W. That鈥檚 not super bright, but it should be plenty of light for routine activities in the tent.

a. Find an expression for the induced current as a function of time if you turn the crank at frequency f. Assume that the semicircle is at its highest point at t=0s.

b. With what frequency will you have to turn the crank for the maximum current to fully light the bulb? Is this feasible?

The current in the solenoid of FIGURE EX30.12 is increasing. The solenoid is surrounded by a conducting loop. Is there a current in the loop? If so, is the loop current cw or ccw?
See all solutions

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