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KCET · Physics · Magnetic Effects of Current

A coil of \(n\) number of turns is wound tightly in the form of a spiral with inner and outer radii a and \(\mathrm{b}\) respectively. When a current of strength \(I\) is passed through the coil, the magnetic field at its centre is

  1. A \(\frac{\mu_{0} n I}{(b-a)} \log _{e} \frac{a}{b}\)
  2. B \(\frac{\mu_{0} n I}{2(b-a)}\)
  3. C \(\frac{2 \mu_{0} \mathrm{nI}}{\mathrm{b}}\)
  4. D \(\frac{\mu_{0} n I}{2(b-a)} \log _{e} \frac{b}{a}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{\mu_{0} n I}{2(b-a)} \log _{e} \frac{b}{a}\)

Step-by-step Solution

Detailed explanation

Consider an element of thickness dr at a distance \(r\) from the centre of spiral coil.

Number of turns in coil \(=\mathrm{n}\)
Number of turns per unit length
\(=\frac{\mathrm{n}}{\mathrm{b}-\mathrm{a}}\)
Number of turns in element \(d r=d n\)
Number of turns per unit length in element \(d r\)
\(=\frac{\mathrm{ndr}}{\mathrm{b}-\mathrm{a}}\)
ie, \(\quad \mathrm{dn}=\frac{\mathrm{ndr}}{\mathrm{b}-\mathrm{a}}\)
Magnetic field at its centre due to element \(d r\) is
\(\mathrm{dB}=\frac{\mu_{0} \mathrm{Idn}}{2 \mathrm{r}}=\frac{\mu_{0} \mathrm{I}}{2} \frac{\mathrm{n}}{(\mathrm{b}-\mathrm{a})} \frac{\mathrm{dr}}{\mathrm{r}} \)
\(\therefore \mathrm{B}=\int_{\mathrm{a}}^{\mathrm{b}} \frac{\mu_{0} \mathrm{Indr}}{2(\mathrm{~b}-\mathrm{a}) \mathrm{r}}=\frac{\mu_{0} \mathrm{In}}{2(\mathrm{~b}-\mathrm{a})} \int_{\mathrm{a}}^{\mathrm{b}} \frac{\mathrm{dr}}{\mathrm{r}} \)
\(=\frac{\mu_{0} \mathrm{In}}{2(\mathrm{~b}-\mathrm{a})} \log _{\mathrm{e}}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)\)