A long insulated copper wire is closely wound as a spiral of $N_{0}$ turn. The spiral lies in the $y - z$ plane and a steady current $I_{0}$ flows through the wire. The X-component of the magnetic field at the centre of the spiral is (assume inner radius as $R_{1}$ and outer radius as $R_{2}$ ).

\frac{μ_{0} N_{0} l_{0}}{4 (R_{2} - R_{1})} l n (R_{2} / R_{1})

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\frac{2 μ_{0} N_{0} l_{0}}{R_{2} - R_{1}} l n (R_{2} / R_{1})

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\frac{μ_{0} N_{0} l_{0}}{2 (R_{2} - R_{1})} l n (R_{2} / R_{1})^{2}

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\frac{μ_{0} N_{0} l_{0}}{2 (R_{2} - R_{1})} l n (R_{2} / R_{1})

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Solution

The correct option is C $\frac{μ_{0} N_{0} l_{0}}{2 (R_{2} - R_{1})} l n (R_{2} / R_{1})$
Consider a small strip of thickness $d r$ at distance $r$ from the centre, then number of turns in this strip would be
$d N = (\frac{N_{0}}{R_{2} - R_{1}}) d r$
Magnetic field due to this element at the centre of the coil will be
$d B = \frac{μ_{0} d N I_{0}}{2 r} = \frac{μ_{0} N_{0} I_{0}}{2 (R_{2} - R_{1})} \frac{d r}{r}$
Total magnetic field at the centre of the spiral
$B = \int_{r = R_{1}}^{r = R_{2}} \frac{μ_{0} N_{0} I_{0}}{2 (R_{2} - R_{1})} \frac{d r}{r} = \frac{μ_{0} N_{0} I_{0}}{2 (R_{2} - R_{1})} l n (\frac{R_{2}}{R_{1}})$ .

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Similar questions

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 :