Understanding Faraday's Law with Solenoids

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The discussion centers on understanding why the electromotive force (emf) in a solenoid is calculated by multiplying the changing magnetic flux for one loop by the number of turns (N). The analogy is made to Ampère's law, where currents add due to the line integral. It is clarified that, similar to adding voltages from batteries in series, each loop in a solenoid contributes to the total emf. For example, if one loop generates 1.5 volts, five loops would produce a total of 7.5 volts. This explanation resolves the initial confusion regarding the addition of turns in the context of Faraday's law.
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It is a well known fact that whenever we want to calculate the emf in a solenoid we usually multiply the changing flux for one loop times N, which is the number of turns in the solenoid.

But why is this?

For example, in the case of amperes law, I know that it makes sense to add currents because you are considering the line integral, one can think of it like

##\oint \mathbf{B} \cdot \mathbf{dl} = \oint (\mathbf{\sum_i B_i}) \cdot \mathbf{dl} = \sum_i ( \oint \mathbf{B} \cdot \mathbf{dl} ) _ i = \sum_i \mu_0 I = \mu_0 \sum_i I_i ##

In that case, currents clearly should add, but I don't see why currents or turns in the solenoid are added in any "deep" sense (when applying Faraday's law).

Thanks and sorry if my question is unclear.
 
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davidbenari said:
It is a well known fact that whenever we want to calculate the emf in a solenoid we usually multiply the changing flux for one loop times N, which is the number of turns in the solenoid.

But why is this?

For example, in the case of amperes law, I know that it makes sense to add currents because you are considering the line integral, one can think of it like

##\oint \mathbf{B} \cdot \mathbf{dl} = \oint (\mathbf{\sum_i B_i}) \cdot \mathbf{dl} = \sum_i ( \oint \mathbf{B} \cdot \mathbf{dl} ) _ i = \sum_i \mu_0 I = \mu_0 \sum_i I_i ##

In that case, currents clearly should add, but I don't see why currents or turns in the solenoid are added in any "deep" sense (when applying Faraday's law).

Thanks and sorry if my question is unclear.
Is there anything you can do to condense this or make it clearer?
 
davidbenari said:
It is a well known fact that whenever we want to calculate the emf in a solenoid we usually multiply the changing flux for one loop times N, which is the number of turns in the solenoid.

But why is this?

Gah, I can't believe nobody tackled this, including me. ?:)

Suppose you have a single loop that produces (for a given changing-magnetic-field configuration) an emf of 1.5 volts. Now suppose you have a coil or solenoid containing, say, 5 of these loops (turns). The loops are electrically in series, so you have 5 emf's in series, 1.5 volts each. It's like having 5 (ideal) dry-cell batteries in series, each with an emf of 1.5 volts, producing a total emf of 7.5 volts.
 
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jtbell:

I guess you have answered my Q. thanks.
 
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