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Magnetism

Section: Physics

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Question : 37 of 46

Marks: +1, -0

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

[JEE Adv 2011 P2]

$\frac{\mu_0 NI}{2(b-a)} \ln(b/a)$
$\frac{\mu_0 NI}{2(b-a)} \ln\left(\frac{b+a}{b-a}\right)$
$\frac{\mu_0 NI}{2} \ln(b/a)$
$\frac{\mu_0 NI}{2b} \ln\left(\frac{b+a}{b-a}\right)$

Solution:

Assume that this geometry of the loop consists of concentric circles whose radii varies from a to b.

The area of the strip is 2π xdx and the current through the area is

\frac{NI}{b-a}

Therefore, the total magnetic field is

B =

\int\limits_{a}^{b} \frac{\mu_0 NI}{2(b-a)x} dx

=

\frac{\mu_0 NI}{2(b-a)} \ln(b/a)

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