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Simple case:-

A bar magnet is inserted into a solenoid with 'n' turns at velocity 'v'. The current is measured using an ammeter that is part of the complete circuit of the solenoid.

My question is that if I increase the number of turns 'n' then will this lead to a greater induced pd and in turn, induced current detected by the ammeter.

Please could you provide a brief explanation as well because its not intuitive whether it will increase or decrease

Shashwat Tomar
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  • Increasing the number of turns also increases the resistance in the circuit. – Farcher May 28 '18 at 17:53
  • Do you really mean 'solenoid', in which the turns are wound helically around a cylinder, or do you mean a flat coil in which all the turns are more of less in one plane? For a solenoid much longer than the magnet, adding more turns to one end, making the solenoid longer still, won't affect the emf, but will reduce the current for the reason given by Farcher. For a flat coil Sagnik Sarkar's answer applies, with Farcher's proviso. – Philip Wood Oct 30 '20 at 17:24
  • Does this answer your question? Axial forces on a solenoid windings – Semoi Feb 28 '21 at 18:12

1 Answers1

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According to Faraday's law of electromagnetism, emf induced is given by

$\epsilon=\frac{d\phi}{dt}$

But $\phi=B.A$

Here B is the electric field and A is the effective area though which the electric flux changes.

$\therefore$effective area =$n.a$

Here a is the area of each turn and n is the number of turns of the coil.

$\therefore \epsilon = \frac{d(B.n.a)}{dt}$

$\implies \epsilon =n.a.\frac{dB}{dt}$

Therefore emf induced is directly proportional to the number of turns in the coil.

By ohm's law, the current in the circuit is proportional to emf across the conductor. Therefore the current, emf both increases with increase in number of turns.

Sagnik Sarkar
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