Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A metal ring 4.50 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic field. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.250 T/s. (a) What is the magnitude of the electric field induced in the ring? (b) In which direction (clockwise or counter clockwise) does the current flow as viewed by someone on the south pole of the magnet?

Short Answer

Expert verified

(a) The magnitude of the electric field induced in the ring is2.81×10-3V/m

(b) The current flows in the counter clockwise direction, as viewed by someone from the south pole of the magnet.

Step by step solution

01

Magnitude of the electric field

Give,

A metal ring with a diameter of d = 4.50cm, magnetic field with magnitude B = 1.12T and the rate of the gradual decrease isdBdt=0.250T

According to the Faraday’s law we get:

ε=E.dl=-dϕBdt

Now, the magnetic flux is :

ϕ=AB=πr2B

Now, integrating the LHS we get:

E.dl=2πr

Therefore,

E2πr=d(Bπr2dt=πr2dBdt=0.0225m20.250T/s=2.81×10-3V/m

Therefore, the electric field is2.81×10-3V/m

02

Direction of the magnetic field

The magnetic field points from north to south and it is decreasing meaning the magnetic flux is decreasing too. Hence the induced magnetic field must be pointed towards the same direction of the original magnetic field. Therefore, the current circulates counter clockwise as viewed from the south pole.

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

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?

You connect a battery, resistor, and capacitor as in Fig. 26.20a, where R = 12.0 Ω and C = 5.00 x 10-6 F. The switch S is closed at t = 0. When the current in the circuit has a magnitude of 3.00 A, the charge on the capacitor is 40.0 x 10-6 C. (a) What is the emf of the battery? (b) At what time t after the switch is closed is the charge on the capacitor equal to 40.0 x 10-6 C? (c) When the current has magnitude 3.00 A, at what rate is energy being (i) stored in the capacitor, (ii) supplied by the battery

A particle with charge-5.60nCis moving in a uniform magnetic fieldrole="math" localid="1655717557369" B=-(1.25T)k^

The magnetic force on the particle is measured to berole="math" localid="1655717706597" F=-(3.40×10-7N)i^-(7.40×10-7N)j^ (a) Calculate all the components of the velocity of the particle that you can from this information. (b) Are there
components of the velocity that are not determined by the measurement of the force? Explain. (c) Calculate the scalar productv֏F. What is the angle between velocity and force?

An 18-gauge copper wire (diameter 1.02 mm) carries a current

with a current density of 3.2×106Am2. The density of free electrons for

copper is8.5×1028electrons per cubic meter. Calculate (a) the current in

the wire and (b) the drift velocity of electrons in the wire.

CALC The region between two concentric conducting spheres with radii and is filled with a conducting material with resistivity ρ. (a) Show that the resistance between the spheres is given by

R=ρ4π(1a-1b)

(b) Derive an expression for the current density as a function of radius, in terms of the potential differenceVab between the spheres. (c) Show that the result in part (a) reduces to Eq. (25.10) when the separation L=b-abetween the spheres is small.
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