A square loop of wire, with sides of length a , lies in the first quadrant of the xy plane, with one comer at the origin. In this region, there is a nonuniform time-dependent magnetic field B(y,t)=ky3t2z^ (where k is a constant). Find the emf induced in the loop.

Short Answer

Expert verified

The induced emf into loop is -kt2a5.

Step by step solution

01

Write the given data from the question.

The side of the square loop is a.

The square loop exists in xy plane.

The non-uniform time dependent magnetic field,B(y,t)=ky3t2z^

02

Calculate the induced emf in the loop.

Let’s take small strip dy on the square loop which at distance y from the x-axis.


The magnetic flux is given by,

ϕ=B(y,t).dA

Here dA is the area of the strip.

Substitute Ky3t2for B(y,t)and ady for dA into above equation.

ϕ=ky3t2.ady

Integrate the above equation to calculate the flus through the entire loop.

ϕ=0aky3t2.adyϕ=kat20ay3dyϕ=kat2y440a

Apply the limits,

ϕ=kat2a44-044ϕ=kat2a44ϕ=kt24a5

The induced emf in the loop is given by,

e=-dϕdt

Substitute kt24a5for ϕinto above equation.

e=-ddtkt24a5e=-2kt4a5e=-kt2a5

Hence the induced emf into loop is -kt2a5.

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 small loop of wire (radius a) is held a distance z above the center of a large loop (radius b ), as shown in Fig. 7.37. The planes of the two loops are parallel, and perpendicular to the common axis.

(a) Suppose current I flows in the big loop. Find the flux through the little loop. (The little loop is so small that you may consider the field of the big loop to be essentially constant.)

(b) Suppose current I flows in the little loop. Find the flux through the big loop. (The little loop is so small that you may treat it as a magnetic dipole.)

(c) Find the mutual inductances, and confirm that M12=M21 ·

A square loop (side a) is mounted on a vertical shaft and rotated at angular velocity ω (Fig. 7.19). A uniform magnetic field B points to the right. Find theεtfor this alternating current generator.

Suppose

E(r,t)=14πε0qr2θ(rυt)r^; B(r,t)=0

(The theta function is defined in Prob. 1.46b). Show that these fields satisfy all of Maxwell's equations, and determine ρ and J. Describe the physical situation that gives rise to these fields.

Try to compute the self-inductance of the "hairpin" loop shown in Fig. 7.38. (Neglect the contribution from the ends; most of the flux comes from the long straight section.) You'll run into a snag that is characteristic of many self-inductance calculations. To get a definite answer, assume the wire has a tiny radius, and ignore any flux through the wire itself.

(a) Referring to Prob. 5.52(a) and Eq. 7.18, show that

E=-At (7.66) for Faraday-induced electric fields. Check this result by taking the divergence and curl of both sides.

(b) A spherical shell of radiusR carries a uniform surface charge σ. It spins about a fixed axis at an angular velocity ω(t)that changes slowly with time. Find the electric field inside and outside the sphere. [Hint: There are two contributions here: the Coulomb field due to the charge, and the Faraday field due to the changing B. Refer to Ex. 5.11.]

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