Question

a coil of 600 turns is threaded by a flux of 8 10^-5 webers. if the flux is reduced to 3 10^-5 webers in 0.015 sec , is average induced e.m.f ?

Answer 1

Answer:

here u go with the answer

Answer 2

Concept:

This problem can be solved by the formula of e.m.f which is Emf = -(dФ/dt).

Given:

Number of turns in coil (N) = 600 turns

Initial Flux = 8 × 10⁻⁵ Wb

Final Flux = 3 × 10⁻⁵ Wb

Time = 0.015 second

Find:

Average induced emf

Solution:

According to the formula

Emf = - N (dФ/dt)

where number of turns are denoted by N, dФ is the change in flux and dt is the time interval.

Change in flux = Final Flux - Initial Flux

Change in flux = (3 × 10⁻⁵) - (8 × 10⁻⁵) Wb

Change in flux = - (5 × 10⁻⁵) Wb

time interval = 0.015 sec

time interval = 15 × 10⁻³ sec

Number of turns in coil = 600

Putting all the values in formula.

Emf = - 600 × (-(5 × 10⁻⁵)/15 × 10⁻³) Volts

Emf = 600 × (5 × 10⁻²/15 ) Volts

Emf = 600 × (10⁻²/3 ) Volts

Emf = 200 × 10⁻² Volts

Emf = 2 Volts

Hence we can say that average induce emf will be 2 volts.

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