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 up through the outer cylinder of radius 2 a . The cylinders

Question

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 up 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 $\frac{3 a}{2}$ from the axis of the cable is

TestHub · Accepted Answer

Using ampere's law for a closed loop of radius $r = \frac{3 a}{2}$ around the axis of the cable,

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

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

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

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

$B = \frac{7 μ_{0} I}{36 π a}$

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