A circular wire loop (radiusr, resistanceR) encloses a region of uniform magnetic field,B, perpendicular to its plane. The field (occupying the shaded region in Fig. 7.56) increases linearly with timeB=αt. An ideal voltmeter (infinite internal resistance) is connected between pointsPandQ.

(a) What is the current in the loop?

(b) What does the voltmeter read? [Answer: αr2/2]

Short Answer

Expert verified

(a)The current in the loop isI=παr2R .

(b) The voltmeter reading isαr22 .

Step by step solution

01

Given information

The radius of circular wire loop is,.

The resistance of circular wire loop is, .

The uniform magnetic field inside the wire loop is, .

The relation between the magnetic field and time is, .

02

Magnetic flux

The magnetic flux inside the wire loop having magnetic field and radius is given by,

03

The current in the loop

(b)

The formula for the emf generated in the loop due to magnetic flux is given by,

Solve further as:

The negative sign indicates the emf value is decreasing.

Also, the emf using Ohm’s law,

Then equating both values,

Hence, the current in the loop is .

04

Determine the voltmeter reading value

(b)

Assume a small elemental region of radius inside the given inside the given region between points P and Q.

For a circle of radius , applying Faraday’s law for a closed area, the formula for the measured emf is given by,

In polar form,

Along the line from P to Q,

and ,

Then the voltage reading between points P and Q can be calculated as,

Hence, the voltmeter reading is .

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

Two coils are wrapped around a cylindrical form in such a way that the same flux passes through every turn of both coils. (In practice this is achieved by inserting an iron core through the cylinder; this has the effect of concentrating the flux.) The primary coil hasN1turns and the secondary hasN2(Fig. 7.57). If the current in the primary is changing, show that the emf in the secondary is given by

ε2ε1=N2N1(7.67)

whereε1is the (back) emf of the primary. [This is a primitive transformer-a device for raising or lowering the emf of an alternating current source. By choosing the appropriate number of turns, any desired secondary emf can be obtained. If you think this violates the conservation of energy, study Prob. 7.58.]

Suppose the circuit in Fig. 7.41 has been connected for a long time when suddenly, at time t=0, switch S is thrown from A to B, bypassing the battery.

Notice the similarity to Eq. 7.28-in a sense, the rectangular toroid is a short coaxial cable, turned on its side.

(a) What is the current at any subsequent time t?

(b) What is the total energy delivered to the resistor?

(c) Show that this is equal to the energy originally stored in the inductor.

Question: A fat wire, radius a, carries a constant current I , uniformly distributed over its cross section. A narrow gap in the wire, of width w << a, forms a parallel-plate capacitor, as shown in Fig. 7.45. Find the magnetic field in the gap, at a distance s < a from the axis.

The magnetic field outside a long straight wire carrying a steady current I is

B=μ02πIsϕ^

The electric field inside the wire is uniform:

E=Iρπa2z^,

Where ρis the resistivity and a is the radius (see Exs. 7.1 and 7 .3). Question: What is the electric field outside the wire? 29 The answer depends on how you complete the circuit. Suppose the current returns along a perfectly conducting grounded coaxial cylinder of radius b (Fig. 7.52). In the region a < s < b, the potential V (s, z) satisfies Laplace's equation, with the boundary conditions

(i) V(a,z)=Iρzπa2 ; (ii) V(b,z)=0

Figure 7.52

This does not suffice to determine the answer-we still need to specify boundary conditions at the two ends (though for a long wire it shouldn't matter much). In the literature, it is customary to sweep this ambiguity under the rug by simply stipulating that V (s,z) is proportional to V (s,z) = zf (s) . On this assumption:

(a) Determine (s).

(b) E (s,z).

(c) Calculate the surface charge density σ(z)on the wire.

[Answer: V=(-Izρ/πa2) This is a peculiar result, since Es and σ(z)are not independent of localid="1658816847863" zas one would certainly expect for a truly infinite wire.]

If a magnetic dipole levitating above an infinite superconducting plane (Pro b. 7 .45) is free to rotate, what orientation will it adopt, and how high above the surface will it float?

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