Does Increasing Coil Turns Decrease Magnetic Flux Density?

[temujin]

Hi

My textbook denotes the magnetic flux through a single turn loop as $$\Phi$$, and the magnetic flux density in the same loop as $$B = \frac{\Phi}{Area}$$.

Extending to an N-turn loop the total flux passing through the coil is given by $$\Psi = N\cdot\Phi \leftrightarrow \Phi = \frac{\Psi}{N}$$ .
Inserted into the equation for B this would produce: $$B=\frac{\Psi}{Area\cdot N}$$

Which means that for a given flux, the flux density would be lower with a high number of turns...!
Can this be right?

Should not the flux density be the total flux passing through the coil divided by the Area of the coil surface...and that increasing N should increase magnetic flux and magnetic flux density...? ?

regards
t.

 

[Jelfish]

I think the problem is that you're confusing your variables. The magnetic flux density of the coil takes into account the magnetic flux due to all of the turns. However, the way you defined phi in your second equation, it seems you're only taking into account one turn of the coil. Instead of psi/N, you shoud have N*phi in your third equation. What you call psi becomes the total flux and what you call phi becomes the flux due to one turn.

 

[thepatientmental]

Hi t,

Thank you for sharing your thoughts on magnetic field density. You are correct that the flux density would be lower with a high number of turns. This is because the total flux remains the same, but it is spread out over a larger area due to the increased number of turns. So while the flux density may decrease, the total flux passing through the coil would remain the same.

To clarify, the flux density is the amount of magnetic flux passing through a unit area. So as the number of turns increases, the area also increases, resulting in a lower flux density. However, as you mentioned, the total flux would also increase as the number of turns increases.

I hope this helps clarify any confusion. Feel free to ask any further questions. Thanks again for your input!