The magnetic field in the cylindrical region shown in figure increases at a constant rate of 20.0 mT/s. Each side of the square loop abcd and defa has a length of 1.00 cm and a resistance of 4.00 Ω. Find the current (magnitude and since) in the wire ad if

(a) the switch S₁ is closed but S₂ is open,

(b) S₁ is open but S₂ is closed,

(c) both S₁ and S₂ are open and

(d) both S₁ and S₂ are closed.

Given:

Rate of increase of magnetic field =

Side length of square loop =

Resistance of each side =4Ω

Area of the coil adef = area of coil abcd =

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 magnetic field is perpendicular to the loop the flux becomes

Rate of change of magnetic field wrt. time is given by

(since area of cross section in magnetic field does not change with time, A remains constant)

Now,

by faraday’s law of electromagnetic induction

…(i)

Where

ϵ =emf produced

ϕ =flux of magnetic field

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

Hence the current through the loop (i) of resistance R is

…(ii)

(a) when the switch S₁ is closed but S₂ is open

no current flows through loop abcd

net resistance of the loop adef R =4× 4 =16Ω

area of loop adef =

using eqn.(ii) current can be given by

As the magnetic field increases, the flux of magnetic field increases in downward direction so by Lenz’s law

The direction of induced current is such that it opposes the change that has induced it

Therefore, current flows in anticlockwise direction (along ad) to increase the magnetic flux in upward direction

(b) S₁ is open but S₂ is closed

No current flows in loop adef

Net resistance of loop abcd=4× 4=16Ω

Area of loop abcd =

using eqn.(ii) current can be given by

As the magnetic field increases, the flux of magnetic field increases in downward direction so by Lenz’s law

Therefore, current flows in anticlockwise direction (along da) to increase the magnetic flux in upward direction.

(c) When both S₁ and S₂ is open

No current flows in both the loop adef and abcd

And hence current in wire ad is zero

(d) When both S₁ and S₂ is closed

The circuit forms a balanced Wheatstone Bridge and the current flowing through the wire ad is zero

Concept of wheat stone bridge:

When the circuit forms a Wheatstone bridge in balanced condition then the current through galvanometer becomes zero