JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Magnetic Effects of Current

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 X-Y plane and a steady current '

I

' flows through the wire. The Z-component of the magnetic field at the center of the spiral is

Answer: (A)

Solution

The magnetic field at the center due to a circular loop of radius r carrying current I is

\frac{μ _{0} I}{2 r}

. For N turns distributed uniformly between radii a and b, the number of turns in a thin shell of thickness dr at radius r is

\frac{N d r}{b - a}

. Integrating:

B = \int_{a}^{b} \frac{μ _{0} I N}{2 r ( b - a )} d r = \frac{μ _{0} I N}{2 ( b - a )} \int_{a}^{b} \frac{d r}{r} = \frac{μ _{0} N I}{2 ( b - a )} ln (\frac{b}{a})

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Single correct option