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.]

Short Answer

Expert verified

The expressionε2ε1=N2N1is verified.

Step by step solution

01

Given information

The number of turns in the primary coils are,N1.

The number of turns in the secondary coils are,N2.

The current passing through the primary coil is,I .

02

Magnetic flux on coils

When the magnitude of the current flowing in the coil varies, a magnetic flux is generated and a certain amount of emf is also induced in the coil.

If the number of turns in the coil are increased then the magnetic flux produced in the coil also increases.

03

Determine the magnetic flux on both coils

When the current passes through a single loop of primary coil, the magnetic field is generated and the change in current value creates the magnetic fluxΦ.

Then, the formula for the magnetic flux on the primary coil is given by,

localid="1658291056948" Φ1=N1Φ

And, the formula for the magnetic flux on the secondary coil is given by,

localid="1658291081661" Φ2=N2Φ

04

Determine emf induced in the primary coil

The formula for the emf induced in the primary coil is given by,

ε1=-dΦ1dtε1=-dN1Φdtε1=-N1dΦdt........(1)

05

Determine emf induced in the secondary coil

The formula for the emf induced in the secondary coil is given by,

ε2=-dΦ2dtε2=-dN2Φdtε2=-N2dΦdt.......(2)

By dividing equation (2) by equation (1),

ε2ε1=-N2dΦdt-N1dΦdtε2ε1=N2N1

Hence proved.

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Most popular questions from this chapter

In the discussion of motional emf (Sect. 7.1.3) Iassumed that the wire loop (Fig. 7.10) has a resistance R; the current generated is then I=vBhR. But what if the wire is made out of perfectly conducting material, so that Ris zero? In that case, the current is limited only by the back emf associated with the self-inductanceL of the loop (which would ordinarily be negligible in comparison with IR). Show that in this regime the loop (massm ) executes simple harmonic motion, and find its frequency. [Answer: ω=Bh/mL].

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.

Two concentric metal spherical shells, of radius a and b, respectively, are separated by weakly conducting material of conductivityσ(Fig. 7 .4a).

(a) If they are maintained at a potential difference V, what current flows from one to the other?

(b) What is the resistance between the shells?

(c) Notice that if b>>a the outer radius (b) is irrelevant. How do you account for that? Exploit this observation to determine the current flowing between two metal spheres, each of radius a, immersed deep in the sea and held quite far apart (Fig. 7 .4b ), if the potential difference between them is V. (This arrangement can be used to measure the conductivity of sea water.)

Question: (a) Use the Neumann formula (Eq. 7.23) to calculate the mutual inductance of the configuration in Fig. 7.37, assuming a is very small (a<<b,a<<z). Compare your answer to Pro b. 7 .22.

(b) For the general case (not assuming is small), show that

M=μ0πβ2abβ(1+158β2+.....)

where

β=abz2+a2+b2

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.

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