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 H

__Answer:-__

Induced emf = (1/2) ω B R^{2}

= (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^{–1}along 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 ^{–3}Ts^{–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 m^{2}

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

Gradient of the magnetic field along negative x-direction,

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

And, rate of decrease of the magnetic field,

(dB/dx) =10^{-3}Ts^{-1}

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

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^{-4}xm^{2}x10^{-1}x0.08

=11.52x10^{-5}Tm^{2}s^{-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^{-5}Tm^{2}s^{-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^{-5}V

∴Induced current, i= (e/R) s

s= (12.96x10^{-5})/ (1.5x10^{-3})

i=2.88x10^{-2}A

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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