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Current is uniformly distributed along the cross section of solid cylinder and radius R. Then the ratio of inductions at points at distances R/2 and 2R from the centre of the cylinder is

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1:2

2:1

1:1

1:4

answer is C.

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Detailed Solution

$B_{I n s i d e} = \frac{μ_{0} i (\frac{R}{2})}{2 π R^{2}}$ $B_{O u t s i d e} = \frac{μ_{0} i}{2 π (2 R)}$

$\frac{B_{I n s i d e}}{B_{o u t s i d e}} = 1$

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