First slide

Magnetic field due to a current carrying circular ring

Question

A coil having N turns is wound tightly in the form of a spiral with inner and outer radii a and b, respectively. When a current I passes through the coil, the magnetic field at the centre is

Difficult

A

$\frac{μ_{0} NI}{b}$
B

$\frac{2 μ_{0} NI}{a}$
C

$\frac{μ_{0} NI}{2 (b - a)} \ln \frac{b}{a}$
D

$\frac{μ_{0} I^{N}}{2 (b - a)} \ln \frac{b}{a}$

Solution

Number of turns per unit width $= \frac{N}{b - a}$ . Consider an elemental ring of radius x and with thickness dx Number of turns in the ring $= dN = \frac{Ndx}{b - a} .$ Magnetic field at the centre due to the ring element
$dB = \frac{μ_{0} (dN) i}{2 x} = \frac{μ_{0} i}{2} \frac{Ndx}{(b - a)} . \frac{1}{x}$
$∴ Field at the centre$
$\begin{array}{l} ∴ Field at the centre \\ = \int dB = \frac{μ_{0} Ni}{2 (b - a)} \int_{a}^{b} \frac{dx}{x} \\ = \frac{μ_{0} Ni}{2 (b - a)} \ln \frac{b}{a} . \end{array}$

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