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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

Question:Calculate the energy stored in the toroidal coil of Ex. 7.11, by applying Eq. 7.35. Use the answer to check Eq. 7.28.

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.

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.

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.]

A transformer (Prob. 7.57) takes an input AC voltage of amplitude V1, and delivers an output voltage of amplitude V2, which is determined by the turns ratio (V2V1=N2N1). If N2>N1, the output voltage is greater than the input voltage. Why doesn't this violate conservation of energy? Answer: Power is the product of voltage and current; if the voltage goes up, the current must come down. The purpose of this problem is to see exactly how this works out, in a simplified model.

(a) In an ideal transformer, the same flux passes through all turns of the primary and of the secondary. Show that in this case M2=L1L2, where Mis the mutual inductance of the coils, and L1,L2, are their individual self-inductances.

(b) Suppose the primary is driven with AC voltage Vin=V1cos(ωt), and the secondary is connected to a resistor, R. Show that the two currents satisfy the relations

L1=dl1dt+Mdl2dt=V1cos(ωt);L1=dl2dt+Mdl1dt=-I2R.

(c) Using the result in (a), solve these equations for localid="1658292112247" l1(t)and l2(t). (Assume l2has no DC component.)

(d) Show that the output voltage (Vout=l2R)divided by the input voltage (Vin)is equal to the turns ratio: VoutVin=N2N1.

(e) Calculate the input power localid="1658292395855" (Pin=Vinl1)and the output power (Pout=Voutl2), and show that their averages over a full cycle are equal.

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