The current in a long solenoid is increasing linearly with time, so the flux is proportional t:.ϕ=αtTwo voltmeters are connected to diametrically opposite points (A and B), together with resistors ( R1and R2), as shown in Fig. 7.55. What is the reading on each voltmeter? Assume that these are ideal voltmeters that draw negligible current (they have huge internal resistance), and that a voltmeter register --abE×dlbetween the terminals and through the meter. [Answer: V1=αR1/(R1+R2). Notice that V1V2, even though they are connected to the same points]

Short Answer

Expert verified

The expression for voltage across the resistance isαR1R1+R2 and is αR2R1+R2.

Step by step solution

01

Write the given data from the question.

The relationship between current and flux,ϕ=αt

The two registers areR1 and R2.

02

Determine the formulas to calculate the voltmeter reading.

The expression for the induced emf is given as follows.

ε=-dϕdt ……. (1)

03

Draw the expression for the voltmeter reading.

Calculate the induced emf.

Substitute αtfor ϕinto equation (1).

ε=ddt(αt)ε=α

The current in the registers is given by

I=εR1+R2

Substituteαforεinto above equation.

I=αR1+R2

The voltage across the registerR1is given by,

V1=IR1

SubstituteαR1+R2forIinto above equation.

V1=αR1+R2R1V1=αR1R1+R2

The voltage across the registerR1is given by,

V2=IR2

Substitute αR1+R2forIinto above equation.

V2=αR1+R2R2V2=αR2R1+R2

Hence, the expression for voltage across the resistanceR1 isαR1R1+R2 and R2is αR2R1+R2.

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Most popular questions from this chapter

A perfectly conducting spherical shell of radius rotates about the z axis with angular velocity ω, in a uniform magnetic field B=B0Z^. Calculate the emf developed between the “north pole” and the equator. Answer:localid="1658295408106" [12B0ωα2].

A long solenoid, of radius a, is driven by an alternating current, so that the field inside is sinusoidal:Bt=B0cosωtz^. 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.

Question: An infinite wire carrying a constant current in the direction is moving in the direction at a constant speed . Find the electric field, in the quasistatic approximation, at the instant the wire coincides with the axis (Fig. 7.54).

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.

Question: The preceding problem was an artificial model for the charging capacitor, designed to avoid complications associated with the current spreading out over the surface of the plates. For a more realistic model, imagine thin wires that connect to the centers of the plates (Fig. 7.46a). Again, the current I is constant, the radius of the capacitor is a, and the separation of the plates is w << a. Assume that the current flows out over the plates in such a way that the surface charge is uniform, at any given time, and is zero at t = 0.

(a) Find the electric field between the plates, as a function of t.

(b) Find the displacement current through a circle of radius in the plane mid-way between the plates. Using this circle as your "Amperian loop," and the flat surface that spans it, find the magnetic field at a distance s from the axis.

Figure 7.46

(c) Repeat part (b), but this time uses the cylindrical surface in Fig. 7.46(b), which is open at the right end and extends to the left through the plate and terminates outside the capacitor. Notice that the displacement current through this surface is zero, and there are two contributions to Ienc.

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