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

Short Answer

Expert verified

The induced emf in the square loop is .Ba2sint

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 .

02

Calculate the generated emf in the square loop.

t=-ddt(BAcost)The area of square loop,A=a2 .

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

tt.

The magnetic flus through the loop is given by,

role="math" localid="1657618600560" =BAcos

Substitute tfor into above equation.

=BAcos

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

(t)=-ddt

Substitute =BAcostfor into above equation.

t=-ddt(BAcost)

Substitute for into above equation.

role="math" localid="1657619049126" t=-ddt(BAcost)t=-Ba2-sint.t=-Ba2sint

Hence the induced emf in the square loop is Ba2sint.

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Most popular questions from this chapter

Prove Alfven's theorem: In a perfectly conducting fluid (say, a gas of free electrons), the magnetic flux through any closed loop moving with the fluid is constant in time. (The magnetic field lines are, as it were, "frozen" into the fluid.)

(a) Use Ohm's law, in the form of Eq. 7.2, together with Faraday's law, to prove that if =and is J finite, then

Bt=(vB)

(b) Let S be the surface bounded by the loop (P)at time t , and S'a surface bounded by the loop in its new position (P')at time t+dt (see Fig. 7.58). The change in flux is

诲桅=S'B(t+dt)da-SB(t)da

Use B=0to show that

S'B(t+dt)da+RB(t+dt)da=SB(t+dt)da

(Where R is the "ribbon" joining P and P' ), and hence that

诲桅=dtSBtda-RB(t+dt)da

(For infinitesimal dt ). Use the method of Sect. 7.1.3 to rewrite the second integral as

dtP(Bv)dI

And invoke Stokes' theorem to conclude that

诲桅dt=S(Bt-vB)da

Together with the result in (a), this proves the theorem.

An infinite wire runs along the z axis; it carries a current I (z) that is a function ofz(but not of t ), and a charge density (t) that is a function of t (but not of z ).

(a) By examining the charge flowing into a segment dz in a time dt, show that d/dt=-di/dz. If we stipulate that (0)=0and I(0)=0, show that (t)=kt, I(z)=-kz, where k is a constant.

(b) Assume for a moment that the process is quasistatic, so the fields are given by Eqs. 2.9 and 5.38. Show that these are in fact the exact fields, by confirming that all four of Maxwell's equations are satisfied. (First do it in differential form, for the region s > 0, then in integral form for the appropriate Gaussian cylinder/Amperian loop straddling the axis.)

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(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 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=t. 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: r2/2]

A long solenoid, of radius a, is driven by an alternating current, so that the field inside is sinusoidal:Bt=B0costz^. A circular loop of wire, of radius a/2 and resistance R , is placed inside the solenoid, and coaxial with it. Find the current induced in the loop, as a function of time.

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