A circular loop of wire with radius r= 0.0480 m and resistance R= 0.160 is in a region of a spatially uniform magnetic field, as shown in Fig. The magnetic field is directed out of the plane of the figure. The magnetic field has an initial value of 8.00 T and is decreasing at a rate of dB/dt= -0.680 T/s.

(a) Is the induced current in the loop clockwise or counterclockwise?

(b) What is the rate at which electrical energy is being dissipated by the resistance of the loop?

Short Answer

Expert verified
  1. The induced current in the loop is in the counterclockwise direction.
  2. The rate at which electrical energy is dissipated by the resistance of the loop is 0.151W.

Step by step solution

01

Concept

The answer to part (a) is given by Lenz's law, which states that the induced current always tends to oppose or cancel out the change that caused it. Our target variable in part (b) is the rate at which energy in the loop is dissipated; Energy is dissipated by the resistance R of the loop at the rate Pdissipated=I2R. The current I in the loop equal /R, where e is the induced emf is given by Faraday's law.

02

Calculate the direction of the inducing current.

By Lenz's law, the induced current must produce a magnetic field Binducedinside the loop that is upward, opposing the change in the flux. By using the right-hand rule for the direction of the magnetic field produced by a circular loop,Binducedwill be in the desired direction if the induced current flows as shown in Fig. given below in the counterclockwise direction.

03

Calculate electric energy.

The magnetic field is uniform over the loop and is perpendicular to the plane of the loop, so the magnetic flux is ϕB=BA, where A=πr2is the area of the circular loop. Hence, by Faraday's law) the induced emf is

ε=-dϕBdt=-ddt(BA)=-AdBdt=-πr2dBdt

So the current in the loop is,

I=|ε|R=πr2RdBdt

Hence,

Pdissipated=I2R=πr2RdBdt2R=π2r4RRdBdt2

Substituting the known values of r, dB/dt, and R, we find

Pdissipated=π(0.0480m)40.160Ω0.680T/s2=0.151mW

Where we used(m2.T)2/s2=(Wb/s)2=V2 for the units.

The electrical energy keeps being dissipated in the loop at the rate calculated in Step 3 until the magnetic flux (or the magnetic field B) stops changing; indeed, substituting dB/dt = 0 into the expression for Pdissipatedgives zero.

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 Vaia!

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

The energy that can be extracted from a storage battery is always less than the energy that goes into it while it is being charged. Why?

Two coils are wound around the same cylindrical form. When the current in the first coil is decreasing at a rate of , the induced emf in the second coil has magnitude 1.65×10-3V. (a) What is the mutual inductance of the pair of coils? (b) If the second coil has 25 turns, what is the flux through each turn when the current in the first coil equals 1.20A? (c) If the current in the second coil increases at a rate of 0.360A/s, what is the magnitude of the induced emf in the first coil?

Batteries are always labeled with their emf; for instances an AA flashlight battery is labelled “ 1.5 V ”. Would it also be appropriate to put a label on batteries starting how much current they provide? Why or why not?

An alpha particle (a He nucleus containing two protons and two neutrons and having a mass of 6.64×10-7kg) travelling horizontally at 35.6km/senter a uniform, vertical,1.80-T magnetic field.(a) What is the diameter of the path followed by this alpha particle? (b) what effect does the magnetic field have on the speed of the particle? (c) What are the magnitude and direction of the acceleration of the alpha particle while it is in the magnetic field? (d) explain why the speed of the particle does not change even though an unbalanced external force acts on it.

When is a 1.5 - VAAA battery not actually a 1.5 - V battery? That is, when do this its terminals provide a potential difference of less than 1.5 V ?

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.

Sign-up for free