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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

21=N2N1(7.67)

where1is 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 expression21=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 flux.

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

localid="1658291056948" 1=N1

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

localid="1658291081661" 2=N2

04

Determine emf induced in the primary coil

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

1=-d1dt1=-dN1dt1=-N1ddt........(1)

05

Determine emf induced in the secondary coil

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

2=-d2dt2=-dN2dt2=-N2ddt.......(2)

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

21=-N2ddt-N1ddt21=N2N1

Hence proved.

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Most popular questions from this chapter

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.

A circular wire loop (radius r , resistance R ) 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 time(B=t)An ideal voltmeter (infinite internal resistance) is connected between points P and Q.

(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.

A perfectly conducting spherical shell of radius rotates about the z axis with angular velocity , in a uniform magnetic field B=B0Z^. Calculate the emf developed between the 鈥渘orth pole鈥 and the equator. Answer:localid="1658295408106" [12B0蝇伪2].

A square loop, side a , resistance R , lies a distance from an infinite straight wire that carries current l (Fig. 7.29). Now someone cuts the wire, so l drops to zero. In what direction does the induced current in the square loop flow, and what total charge passes a given point in the loop during the time this current flows? If you don't like the scissors model, turn the current down gradually:

I(t)={(1-t)I0for0t1/afort>/a

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