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. 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.

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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