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

A long solenoid with radius a and n turns per unit length carries a time-dependent currentl(t) in theϕ^ direction. Find the electric field (magnitude and direction) at a distance s from the axis (both inside and outside the solenoid), in the quasistatic approximation.

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) Show that Maxwell's equations with magnetic charge (Eq. 7.44) are invariant under the duality transformation

E'=Ecosα+cBsinα,cB'=cBcosα-Esinα,cq'e=cqecosα+qmsinα,q'm=qmcosα-cqesinα,

Where c1/ε0μ0and αis an arbitrary rotation angle in “E/B-space.” Charge and current densities transform in the same way as qeand qm. [This means, in particular, that if you know the fields produced by a configuration of electric charge, you can immediately (using α=90°) write down the fields produced by the corresponding arrangement of magnetic charge.]

(b) Show that the force law (Prob. 7.38)

F=qe(E+V×B)+qm(B-1c2V×E)

is also invariant under the duality transformation.

Find the self-inductance per unit length of a long solenoid, of radius R , carrying n turns per unit length.

As a lecture demonstration a short cylindrical bar magnet is dropped down a vertical aluminum pipe of slightly larger diameter, about 2 meters long. It takes several seconds to emerge at the bottom, whereas an otherwise identical piece of unmagnetized iron makes the trip in a fraction of a second. Explain why the magnet falls more slowly.

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