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Chapter 30: Q.47 (page 872)

FIGURE P30.47 shows a 1.0-cm-diameter loop with R=0.50Ωinside a 2.0-cm-diameter solenoid. The solenoid is 8.0cm long, has 120 turns, and carries the current shown in the graph. A positive current is CWwhen seen from the left. Determine the current in the loop att=0.010s

Short Answer

Expert verified

The current induced in the loop is −15μA

Step by step solution

01

 Lenz's Law

Inside this illustration here, we've shown the approach as from left. This power supplied in the thread will create a magnetic field that resisted the variation due to Lenz's law. In our situation, the stream with in circuit will turn around to achieve this need.

02

Step 2:Current in the loop

Just to see if can have discover its power in the spiral. The solenoid's magnetic field is consistent, and so its value is calculated

B=μ0IsolNl​A=14d2π

Φ=BA=14μ0IsolNld2π

Faraday's law of induction

ε=−ΔΦΔt

The magnetic flux is the current

ΔΦΔt=14lμ0Nd2πΔIsolΔt

03

Ohm's Law

The loop of current using Ohm's law

Iloop=εR=−14lRμ0Nd2πΔIsolΔt.

The graph depicts the change of piezo power - the charging current. by

ΔIsolΔt=0.5A0.01s=50A/s

Iloop=−15μA

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Most popular questions from this chapter

An 8.0cm×8.0cmsquare loop is halfway into a magnetic CALC field perpendicular to the plane of the loop. The loop's mass is and its resistance is 0.010Ω. A switch is closed at t=0s, causing the magnetic field to increase from0to1.0Tin0.010s .

a. What is the induced current in the square loop?

b. With what speed is the loop "kicked" away from the magnetic field?

Hint: What is the impulse on the loop?

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≤t≤0.04sfor the case thatlocalid="1648816410574" width="87">v0=10m/s,localid="1648921234041" l=10cm,localid="1648921244487" m=1.0g,localid="1648921254639" R=0.0010Ω,and localid="1648921264943" B=0.10T. The back edge of the loop does not reach the field during this time interval.

The switch in FIGURE EX30.33 has been in position 1 for a long time. It is changed to position 2 at t = 0 s. a. What is the maximum current through the inductor? b. What is the first time at which the current is maximum?

20. The magnetic field inside a 5.0-cm-diameter solenoid is2.0T and decreasing at 4.0T/s. What is the electric field strength inside the solenoid at a point (a) on the axis and (b) 2.0cmfrom the axis?

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 1.0Ωlightbulb rated at 4.0W. That’s 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?
See all solutions

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