The magnetic field Bin a certain region is 0.128 T, and its direction is that of the +z-axis in below figure (a) What is the magnetic flux across the surface in the figure? (b) What is the magnetic flux across the surface aefd? (c) What is the magnetic flux across the surface befc? (d) What the net flux through all five surface that enclose the shaded volume.

Short Answer

Expert verified

(a) The magnetic flux across the surface isfabcd=0Wb

(b) The magnetic flux across the surface is fbefc=-0.0115Wb.

(c) The magnetic flux across the surface is faefd=+0.0115Wb.

(d) The net flux through all five surface that enclose the shaded volume is 0Wb.

Step by step solution

01

Definition of magnetic flux

The term magnetic flux may be defined as the measurement of total magnetic field passing through the given area.

02

Determine the magnetic flux of surface area abcd and befc

The magnetic flux can be calculated as

From figure see that area vectors are perpendicular to the surface.

fabcd=B×Afabcd=BACos90fabcd=0

Hence, the magnetic flux across the surface isfabcd=0Wb

The area of the surface in the opposite direction of magnetic field so the magnetic flux calculated as

fbefo=B×Afbefo=BACos180fbeto=BAfbefc=B(be)(ef)

Here

be=0.300mef=0.300m

than

fbeft=(0.128T)(0.300m)(0.300m)fbeft=0.0115Wb

Hence, the magnetic flux across the surface is fbefc=-0.0115Wb.

03

Determine the magnetic flux of surface area and net flux

The angle between surface area and the magnetic field is

cos(θ)=30.050.0cos(θ)=35

Now the magnetic flux is

faetd=B×Afaefd=3BA5faefd=35(0.128T)(0.500m)(0.300m)faefd=+0.0115Wb

Hence, the magnetic flux across the surface aefd is faefd=+0.0115Wb.

And the net flux is zero because angle between them is90° .than

ftotal=fbefic+faefdftobl=0.0115Wb+0.0115Wbftotal=0

Hence, the net flux through all five surface that enclose the shaded volume is 0Wb.

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

A typical small flashlight contains two batteries, each having an emf of1.5V, connected in series with a bulb having resistance17Ω. (a) If the internal resistance of the batteries is negligible, what power is delivered to the bulb? (b) If the batteries last for1.5hwhat is the total energy delivered to the bulb? (c) The resistance of real batteries increases as they run down. If the initial internal resistance is negligible, what is the combined internal resistance of both batteries when the power to the bulb has decreased to half its initial value? (Assume that the resistance of the bulb is constant. Actually, it will change somewhat when the current through the filament changes, because this changes the temperature of the filament and hence the resistivity of the filament wire.)

We have seen that a coulomb is an enormous amount of charge; it is virtually impossible to place a charge of 1 C on an object. Yet, a current of 10A,10C/sis quite reasonable. Explain this apparent discrepancy.

The definition of resistivity (ρ=EJ) implies that an electrical field exist inside a conductor. Yet we saw that in chapter 21 there can be no electrostatic electric field inside a conductor. Is there can be contradiction here? Explain.

In a cyclotron, the orbital radius of protons with energy 300keVis 16.0cm. You are redesigning the cyclotron to be used instead for alpha particles with energy 300keV. An alpha particle has chargeq=+2e and mass m=6.64×10-27kg. If the magnetic filed isn't changed, what will be the orbital radius of the alpha particles?

Section 27.2 describes a procedure for finding the direction of the magnetic force using your right hand. If you use the same procedure, but with your left hand, will you get the correct direction for the force? Explain.

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