Discussion Forum



1)

A long insulated copper wire is closely wound as a spiral of N turns. The spiral has inner radius a and outer radius b. The spiral lies in the XY-plane and a steady current I flows through the wire. The Z-component of the magnetic field at the centre of the spiral is

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A) $\frac{\mu_{0}Nl}{2(b-a)} ln \left(\frac{b}{a}\right)$

B) $\frac{\mu_{0}Nl}{2(b-a)} ln \left(\frac{b+a}{b-a}\right)$

C) $\frac{\mu_{0}Nl}{2b} ln \left(\frac{b}{a}\right)$

D) $\frac{\mu_{0}Nl}{2b} ln \left(\frac{b+a}{b-a}\right)$

Answer:

Option A

Explanation:

If we takes small strip of dr at a distance r from the centre, then the number of turns in this strip would be

 $dN= \frac{N}{b-a} dr$

 Magnetic field due to this element at the centre of the coil will be

 dB= $\frac{ \mu_{0}(dN)I}{2r}=\frac{ \mu_{0} NI}{2(b-a)} \frac{dr}{r}$

 $\therefore$  $B=\int_{r=a}^{r=b} dB=\frac{\mu_{0}NI}{2(b-a)}ln \frac{b}{a}$

$\therefore$  correct answer is (a)