Due to a small magnet intensity at a distance `x` in the end on position is `9` Gauss. What will be the intensity at a distance `(x)/(2)` on broad side on position?

Options

A. `9` Gauss
B. `4` Gauss
C. `36` Gauss
D. `4.5` Gauss

Answer

C

Explanation

In C.G.S. `B_("axial")=9=(2M)/x^(3)` …(i)
`B_("equaterial")=M/(x/2)^(3)=(8M)/x^(3)` …(ii)
From equations (i) and (ii) `B_("equaterial")=36` Gauss