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A coil having $N$ turns is wound tightly in the form of a spiral with inner and outer radii as $a$ and $b$ respectively. The magnetic field at the centre when a current $i$ passes through it is

$\begin {array} {1 1} (a)\;\large\frac{\mu_0Ni}{a} & \quad (b)\;\large\frac{\mu_0Ni}{b} \\ (c)\;\large\frac{\mu_0Ni}{2(b-a)}In \bigg( \large\frac{b}{a} \bigg) & \quad (d)\;\large\frac{\mu_0Ni}{(b-a)}In \bigg( \large\frac{b}{a} \bigg) \end {array}$


How did u write the second line for dN

1 Answer

Consider an element of thickness dr at a distance r from the centre.
The number of turns in this element is $dN =\large\frac{N}{(b-a)}$$dr$
Magnetic field at the centre of the coil due to this element is $dB = \large\frac{\mu_0(dn)i}{2r}$
Integrating from a to b gives the desired answer
Ans : (c)
answered Feb 19, 2014 by thanvigandhi_1

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