The figure shows a long coaxial cable in which a current $I$ flows down through the inner cylinder of radius $a$ and the same current flows back through the outer cylinder of radius $2 a$ . The cylinders are insulated from each other and the current is uniformly distributed over the area of the cross-section in each cylinder. The strength of the magnetic field at a distance $3 a / 2$ from the axis of the cable is $B = \frac{7 μ_{0} I}{n π a}$ the value of $n$ is___________

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answer is 36.

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Detailed Solution

Using ampere's law for a closed loop of radius r = 3a/2 round the axis of the cable.

$\oint \vec{B} . d \vec{l} . = μ_{0} [I - \frac{I (π r^{2} - π a^{2})}{(4 π a^{2} - π a^{2})}]$

$B \times 2 π r = μ_{0} [I - \frac{I (π r^{2} - π a^{2})}{(4 π a^{2} - π a^{2})}]$

$B \times 2 π \frac{3 a}{2} = μ_{0} [\frac{I - (π \frac{9 a^{2}}{4} - π a^{2})}{3 π a^{2}}]$

$B \times 3 π a = μ_{0} [I - \frac{5 I}{12}] = \frac{7 μ_{0} I}{36 π a}$

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