Q. 204.5( 4 Votes )

# Figure shows a circular coil of N turns and radius a, connected to a battery of emf ϵ through a rheostat. The rheostat has a total length L and resistance R. The resistance of the coil is r. A small circular loop of radius a’ and resistance r’ is placed coaxially with the coil. The center of the loop is at a distance x from the center of the coil. In the beginning, the sliding contact of the rheostat is at the left end and then onwards it is moved towards right at a constant speed v. Find the emf induced in the small circular loop at the instant

(a) the contact begins to slide and

(b) it has slid through half the length of the rheostat.

Answer :

Given:

Area of coil (2) of radius a’ =

We know that magnetic field due to coil (1) at the center of coil (2) is

Where

N=no. of turns in coil (1)

i= current in coil (1)

a=radius of coil (1)

x=distance of center of coil (2) from center of coil (1)

We know that,

Flux (ϕ) of magnetic field (B) through the loop of cross section area A in the magnetic field is given by

Since the magnetic field due to coil (1) is parallel to axis of coil (2) θ =0° and flux through the coil (2) is given by

Now,

by faraday’s law of electromagnetic induction

…(i)

Where

e =emf produced

ϕ =flux of magnetic field

using eqn.(i) emf induced in the coil (2) is given by

. (ii)

Let y be the distance of sliding contact from its right end

Given,

Total length of rheostat =L

Total resistance of rheostat=R

When the sliding contact is at a distance y from its right end then the resistance (R’) of the rheostat is given by

So the current i flowing through the circuit is given by

Where r is the resistance of the coil and ϵ is the emf of battery

Putting value of i in eqn.(ii) we get,

Since

(a) When the contact begins to slide

Therefore, magnitude of emf induced is

(b) When the contact has slid through half the length of rheostat

Therefore, magnitude of emf induced is

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