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 [2011]

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Q.
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 [2011]

1 $\frac{μ_{0} N I}{2 (b - a)} \ln (\frac{b}{a})$

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

3 $\frac{μ_{0} N I}{2 b} \ln (\frac{b}{a})$

4 $\frac{μ_{0} N I}{2 b} \ln (\frac{b + a}{b - a})$

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Solution

Q. 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 [2011]

1 μ0NI2(b-a)ln(ba)

2 μ0NI2(b-a)ln(b+a b-a )

3 μ0NI2bln(ba)

4 μ0NI2bln(b+a b-a )

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