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There's a formula for self inductance: $$ L=\frac{n\Phi}{i}$$ where n is the number of loops.

But the book also says self inductance is directly proportional to $ n^2 $

I totally agree with the second sentence. But the formula seems to show $L$ is directly proprtional to $n $ and not $ n^2 $.

I am confused. Can you help?

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2 Answers 2

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Notice, the magnetic magnetic field $B$ at the center of a coil carrying current $i$, with radius $r$ & having $n$ no. of turns $$B=\frac{\mu_0}{2}\frac{ni}{r}$$ hence, magnetic flux $\phi$ linked to the coil is given as $$\Phi=BA=\frac{\mu_0}{2}\frac{ni}{r}\pi r^2=\frac{\mu_0 \pi nir}{2}$$ Now, setting the value of $\phi$, we get $$L=\frac{n\Phi}{i}=\frac{n\frac{\mu_0 \pi nir}{2}}{i}=\frac{\mu_0 \pi n^2r}{2}$$ $$L\propto n^2$$ It is obvious that keeping other parameters constant, the self inductance $\color{red}{L}$ of a coil is directly proportional to $\color{red}{n^2}$

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    $\begingroup$ Oh, forgot to see the n hiding inside flux... thanks $\endgroup$
    – Shubham
    Commented Nov 26, 2015 at 7:55
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    $\begingroup$ You're also in the Physics community... congratulations on your astuteness. $\endgroup$
    – Sebastiano
    Commented Jul 19, 2020 at 23:24
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From your question,

Self inductance $L$ : $$L=\frac{n\phi}{i}$$ Magnetic flux, is the product of Magnetic field and the area of cross section intercepted by magnetic field lines $$\phi=BA$$

Magnetic field is directly proportional to the number of turns in the inductor $$B\propto n$$

$$\phi\propto n$$ $$L\propto n\cdot n$$ $$\therefore L\propto n^2$$ I hope it helps you

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