A long co-axial cable consists of a solid cylindrical conductor of radius $R_{1} = R$ surrounded by a thin conducting cylindrical shell of radius $R_{2} = 2 R,$ The inner cylinder carries a current $I_{1} = I$ and the outer cylindrical shell carries a current $I_{2} = \frac{I}{2},$ in the same direction. What is the magnitude of the magnetic field at point which is at a distance of $3 R$ from the centre?

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$\frac{μ_{0} I}{9 πR}$

$\frac{μ_{0} I}{6 πR}$

$\frac{μ_{0} I}{3 πR}$

$\frac{μ_{0} I}{4 πR}$

answer is D.

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Detailed Solution

In this case:

$B \times 2 πr = μ_{0} i = μ_{0} (I_{1} + I_{2}) B = \frac{μ_{0} (I_{1} + I_{2})}{2 πr}$

Putting $I_{1} = I$ , $I_{2} = \frac{I}{2}$ and $r = 3 R$ we get

$B = \frac{μ_{0} I}{4 πR}$

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