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

Short Answer

Expert verified

The induced emf in the square loop is .Ba2Ó¬sinÓ¬t

Step by step solution

01

Write the given data from the question.

The uniform magnetic field is B .

The side of the square loop is a .

The angular velocity is Ó¬.

02

Calculate the generated emf in the square loop.

εt=-ddt(BAcosӬt)The area of square loop,A=a2 .

The square loop is moving at angle with angular velocity in time

tθӬt.

The magnetic flus through the loop is given by,

role="math" localid="1657618600560" ϕ=BAcosθ

Substitute Ӭtfor θinto above equation.

ϕ=BAcosθ

According to the Faraday’s law, the induced emf in any closed loop is equal to the negative of the rate of change of flux in the circuit.

ε(t)=-dϕdt

Substitute ϕ=BAcosӬtfor ϕinto above equation.

εt=-ddt(BAcosӬt)

Substitute for into above equation.

role="math" localid="1657619049126" εt=-ddt(BAcosӬt)εt=-Ba2-sinӬt.Ӭεt=-Ba2sinӬt

Hence the induced emf in the square loop is Ba2Ó¬sinÓ¬t.

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Most popular questions from this chapter

A familiar demonstration of superconductivity (Prob. 7.44) is the levitation of a magnet over a piece of superconducting material. This phenomenon can be analyzed using the method of images. Treat the magnet as a perfect dipole , m a height z above the origin (and constrained to point in the z direction), and pretend that the superconductor occupies the entire half-space below the xy plane. Because of the Meissner effect, B = 0 for Z≤0, and since B is divergenceless, the normal ( z) component is continuous, so Bz=0just above the surface. This boundary condition is met by the image configuration in which an identical dipole is placed at - z , as a stand-in for the superconductor; the two arrangements therefore produce the same magnetic field in the region z>0.

(a) Which way should the image dipole point (+ z or -z)?

(b) Find the force on the magnet due to the induced currents in the superconductor (which is to say, the force due to the image dipole). Set it equal to Mg (where M is the mass of the magnet) to determine the height h at which the magnet will "float." [Hint: Refer to Prob. 6.3.]

(c) The induced current on the surface of the superconductor ( xy the plane) can be determined from the boundary condition on the tangential component of B (Eq. 5.76): B=μ0(K×z^). Using the field you get from the image configuration, show that

K=-3mrh2Ï€(r2+h2)52Ï•^

where r is the distance from the origin.

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.

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 of wire (side a) lies on a table, a distance s from a very long straight wire, which carries a current I, as shown in Fig. 7.18.

(a) Find the flux of B through the loop.

(b) If someone now pulls the loop directly away from the wire, at speed, V what emf is generated? In what direction (clockwise or counter clockwise) does the current flow?

(c) What if the loop is pulled to the right at speed V ?

Refer to Prob. 7.16, to which the correct answer was

E(s,t)=μ0I0Ӭ2ττsin(Ӭt)In(as)z^

(a) Find the displacement current density Jd·

(b) Integrate it to get the total displacement current,

Id=∫Jd.da

Compare Id and I. (What's their ratio?) If the outer cylinder were, say, 2 mm in diameter, how high would the frequency have to be, forId to be 1% of I ? [This problem is designed to indicate why Faraday never discovered displacement currents, and why it is ordinarily safe to ignore them unless the frequency is extremely high.]

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