A coil of n number of turns wound tightly in the form of a spiral with inner and outer radii r1 and r2 respectively. When a current of strength I is passed through the coil, the magnetic field at its centre will be :

\mu _0nI \over 2(r_2 - r_1)\log _er_2 \over r_1

\mu _0nI \over r_2 - r_1\log _er_1 \over r_2

Magnetics PYQ — JEE Main 2022 · 30 Jun · Shift 1 · Q50 with Solution

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Correct answer: D

For a circular loop of radius $r$ carrying current $I$ , the magnetic field at the centre is

$B=\frac{\mu_0 I}{2r}$

The coil has $n$ turns tightly wound between inner radius $r_1$ and outer radius $r_2$ .

Since the winding is tight, we can treat the turns as continuously distributed in radius.

So the number of turns per unit radial length is

$\frac{dn}{dr}=\frac{n}{r_2-r_1}$

Hence,

$dn=\frac{n}{r_2-r_1}\,dr$

A small group of turns $dn$ at radius $r$ contributes field

$dB=\frac{\mu_0 I}{2r}\,dn$

Substitute $dn$ :

$dB=\frac{\mu_0 I}{2r}\cdot \frac{n}{r_2-r_1}\,dr$

=\frac{\mu_0 n I}{2(r_2-r_1)}\int_{r_1}^{r_2}\frac{dr}{r}$$ Using $$\int \frac{dr}{r}=\ln r$$ we get $$B=\frac{\mu_0 n I}{2(r_2-r_1)}\left[\ln r\right]_{r_1}^{r_2}$$ $$B=\frac{\mu_0 n I}{2(r_2-r_1)}\ln\left(\frac{r_2}{r_1}\right)$$ 5. **Match with the options** This is exactly $$\boxed{\frac{\mu_0 n I}{2(r_2-r_1)}\ln\left(\frac{r_2}{r_1}\right)}$$ So the correct option is **D**.