Consider a coaxial cable which consists of an inner wire of radius $a$ surrounded by an outer cylindrical shell of inner and outer radii $b$ and $c$ respectively. The inner wire carries an electric current $i_{0}$ and the outer shell carries an equal current in opposite direction. Then, the magnetic field at a distance x from the axis is…… (Assume that the current density is uniform in the inner wire and also uniform in the outer shell).

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$\frac{μ_{0} i_{0} x}{2 π a^{2}} f o r x < a$

$\frac{μ_{0} i_{0}}{2 π x} f o r a < x < b$

$\frac{μ_{0} i_{0} (c^{2} - x^{2})}{2 π x (c^{2} - b^{2})} f o r b < x < c$

$B = 0 f o r x > c$

answer is A, B, C, D.

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Detailed Solution

Use Ampere's circuital law

For x < a, $B (2 π x) = μ_{0} \frac{i_{0}}{π a^{2}} π x^{2} \Rightarrow B = \frac{μ_{0} i_{0}}{2 π a^{2}} x$ . Option (a) is correct.
For a < x < b, $B (2 π x) = μ_{0} i_{0} \Rightarrow B = \frac{μ_{0} i_{0}}{2 π x}$ Option (b) is correct.
For b < x < c, $B (2 π x) = μ_{0} [i_{0} - \frac{i_{0}}{π (c^{2} - b^{2})} π (x^{2} - b^{2})] \Rightarrow B = \frac{μ_{0} i_{0}}{2 π x} (\frac{c^{2} - x^{2}}{c^{2} - b^{2}})$
Option (c) is correct.
B = 0 for x > c because current threading the loop is zero. Option (d) is correct.