/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q80P In Fig. 30-63, R=4.0°ìΩ , L=8.... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Fig. 30-63, R=4.0°ìΩ,L=8.0μ±áand the ideal battery hasε=20v. How long after switch S is closed is the current 2.0 mA?

Short Answer

Expert verified

For 1.0×10-9safter switch S is closed, the value for the current is 2.0 mA

Step by step solution

01

Given

R=4.0°ìΩ=4.0×103ΩL=8.0μ±á=8.0×10-6Hε=20vI=2.0mA=2.0×10-3A

02

Understanding the concept

We have the equation for current in the loop. We rearrange the equation to find the required value.

Formula:

I=εR1-e-tτL

τL=LR

03

Calculate how long after switch S is closed is the current 2.0 mA

We have,

I=εR1-e-tτL

But,

τL=LRτL=8.0×10-64.0×103τL=2.0×10-9s2.0×103=204.0×103×1-e-t2.0×10-91-e-t2.0×10-9=2.0×10-3×4.0×103201-e-t2.0×10-9=0.4e-t2.0×10-9=0.4-1e-t2.0×10-9=-0.6e-t2.0×10-9=0.6

Taking natural log at both the sides, we get

-t2.0×10-9=In(0.6)t2.0×10-9=0.51t=2.0×10-9×0.51t=1.0×10-9s

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

As seen in Figure, a square loop of wire has sides of length 2.0 cm. A magnetic field is directed out of the page; its magnitude is given by B=4.0t2y, where B is in Tesla, t is in seconds, and y is in meters. At t = 2.5 s, (a) what is the magnitude of the emf induced in the loop? (b) what is the direction of the emf induced in the loop?

In Figure, a wire forms a closed circular loop, of radius R = 2.0mand resistance4.0Ω. The circle is centered on a long straight wire; at time t = 0, the current in the long straight wire is 5.0 Arightward. Thereafter, the current changes according toi=5.0A-(2.0As2)t2. (The straight wire is insulated; so there is no electrical contact between it and the wire of the loop.) What is the magnitude of the current induced in the loop at times t > 0?

A coil with an inductance of2.0 H and a resistance of 10Ωis suddenly connected to an ideal battery with ε=100V. At 0.10 safter the connection is made, (a) what is the rate at which energy is being stored in the magnetic field? (b) what is the rate at which thermal energy is appearing in the resistance? (c) what is the rate at which energy is being delivered by the battery?

A rectangular coil of N turns and of length a and width b is rotated at frequency f in a uniform magnetic field, as indicated in Figure. The coil is connected to co-rotating cylinders, against which metal brushes slide to make contact. (a) Show that the emf induced in the coil is given (as a function of time t) byε=2ττfNabsin(2ττft)=ε0sin(2ττft). This is the principle of the commercial alternating-current generator. (b) What value of Nabgives an emf withε0150Vwhen the loop is rotated at 60.0revs in a uniform magnetic field of 0.500 T?

One hundred turns of (insulated) copper wire are wrapped around a wooden cylindrical core of cross-sectional area 1.20×10-3m2. The two ends of the wire are connected to a resistor. The total resistance in the circuit is13.0Ω. If an externally applied uniform longitudinal magnetic field in the core changes from 1.60 Tin one direction to1.60 T in the opposite direction, how much charge flows through a point in the circuit during the change?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.