A coil of resistance 1000 is placed in a magnetic field of 1 m Wb. The coil has 100 turns and a galvanometer of 400Ω resistance is connected in series with it: find the average e.m.f. and the current, if the coil is moved in 1/10th s from the given field to a field of 0.2 m Wb. (Answer: 1.6 mA ). 



Data Given:

Resistance of coil = R₁ = 400 Ω

Initial magnetic flux = ф₁ = 1 m Wb = 1 ｘ 10⁻³ Wb

Numbers of turns = N = 100 turns

Resistance of Galvanometer = R₂ = 400 Ω

Final magnetic flux = ф₂ = 0.2 m Wb = 0.2 ｘ 10⁻³ Wb

Change in time interval = Δt = `\frac{1}{10}` s = 0.1 s



To Find:

Average e.m.f. = Ɛ = ?

Average current = I = ?

Solution:

According to Faraday's Law

Ɛ = - N `\frac {Δф}{Δt}`

where Δф = ф₂ - ф₁ = 0.2 ｘ 10⁻³ Wb - 1 ｘ 10⁻³ Wb = - 0.8 ｘ 10⁻³ Wb

by putting the corresponding values

Ɛ = - 100 ｘ `\frac {- 0.8 ｘ 10⁻³ Wb}{0.1 s}`

Ɛ =  100 ｘ 8 ｘ 10⁻³ Web s⁻¹

Ɛ =  0.8 V ---------------Ans.(1)


To find the average current I we are using Ohm's law

V = R I

or 

I = `\frac {V}{R}`  -------------(1)


As the Galvanometer and the coil is connected in series. So  

`\R_{eq}` =  R₁ + R₂ = 100 Ω + 400 Ω = 500 Ω

Now equation (1)

I = 
`\frac {0.8 V}{500 Ω}`

I = 0.0016 A

0r

I = 1.6 mA ---------------------------Ans. (2)




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