Class 12 Physics Electromagnetic Induction Varying Area A in a closed loop with movable piston

Methods of producing Induced Emf in a circuit: Varying Area (A) in a closed loop(with movable piston)

  • Consider a closed loop such that one arm of the loop acts as a movable piston.
  • The loop is present inside a magnetic field.
  • The movement of piston will cause a change in area of the loop.
  • As there is change in area as aresult there is change in flux because of which there is induced emf.

 

 

Problem:- A wheel with 10 metallic spokes each 0.5 m long is rotated with aspeed of 120 rev/min in a plane normal to the horizontal componentof earth’s magnetic field HE at a place. If HE = 0.4 G at the place, whatis the induced Emf between the axle and the rim of the wheel?  Note that 1 G = 10–4 T.

Answer:-

Induced emf = (1/2) ω B R2

= (1/2) × 4π × 0.4 × 10–4 × (0.5)2

= 6.28 × 10–5 V

The number of spokes is immaterial because the emf across thespokes is in parallel.

Problem:-

A square loop of side 12 cm with its sides parallel to X and Y axes ismoved with a velocity of 8 cm s–1 in the positive x-direction in anenvironment containing a magnetic field in the positive z-direction.The field is neither uniform in space nor constant in time. It has a

gradient of 10 –3 T cm–1along the negative x-direction (that is it increasesby 10 – 3 T cm–1

as one moves in the negative x-direction), and it isdecreasing in time at the rate of

10 –3Ts–1. Determine the direction andmagnitude of the induced current in the loop if its resistance is 4.50 mΩ.

Answer:-

Side of the square loop, s = 12 cm = 0.12 m

Area of the square loop, A = 0.12 × 0.12 = 0.0144 m2

Velocity of the loop, v = 8 cm/s = 0.08 m/s

Gradient of the magnetic field along negative x-direction,

(dB/dx) =10-3Tcm-1 =10-1Tm-1

And, rate of decrease of the magnetic field,

(dB/dx) =10-3Ts-1

Resistance of the loop, R=4.5 mΩ=4.5x10-3Ω

Rate of change of the magnetic flux due to the motion of the loop in a non-uniform magnetic field is given as:

(dΦ/dt) =A x (dB/dt) xv

=144x 10-4xm2x10-1x0.08

=11.52x10-5Tm2s-1

Rate of change of the flux due to explicit time variation in field B is given as:

(dΦ/dt) =A x (dB/dt)

=144x10-5Tm2s-1

Since the rate of change of the flux is the induced Emf, the total induced Emf in the loop can be calculated as:

e=144x10-5+11.52x10-5

=12.96 x 10-5V

∴Induced current, i= (e/R) s

s= (12.96x10-5)/ (1.5x10-3)

i=2.88x10-2A

Hence, the direction of the induced current is such that there is an increase in the flux through the loop along positive z-direction.

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