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

Question

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 (A) (μ0 NI /2( b - a )) ln (( b / a )) (B) (μ0 NI /2( b - a )) ln (( b + a / b - a )) (C) (μ0 NI /2 b ) ln (( b / a )) (D) (μ0 NI /2 b ) ln (( b + a / b - a ))

Tardigrade Admin · Accepted Answer

∫ limitsab (μ0 IN /2 r(b-a)) dr =(μ0 I N/2(b-a)) ∫ limitsab (d r/r)=(μ0[N/2(b-a)) ln ((b/a))

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

$\frac{μ _{0} N I}{2 ( b - a )} ln (\frac{b}{a})$

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

$\frac{μ _{0} N I}{2 b} ln (\frac{b}{a})$

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

$a \int b \frac{μ _{0} I N}{2 r ( b - a )} d r$
$= \frac{μ _{0} I N}{2 ( b - a )} a \int b \frac{d r}{r} = \frac{μ _{0} [ N}{2 ( b - a )} ln (\frac{b}{a})$

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

2(b−a)μ0​NI​ln(ab​)

2(b−a)μ0​NI​ln(b−ab+a​)

2bμ0​NI​ln(ab​)

2bμ0​NI​ln(b−ab+a​)