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5.Magnetism and Matter
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The magnetic field at a point $x $ on the axis of a small bar magnet is equal to the field at a point $ y $ on the equator of the same magnet. The ratio of the distances of $x$ and $y$ from the centre of the magnet is
A
${2^{ - 3}}$
B
${2^{ - 1/3}}$
C
${2^3}$
D
${2^{1/3}}$
Solution
(d)${B_1} = \frac{{2M}}{{{x^3}}}\;and\;{B_2} = \frac{M}{{{y^3}}}$
As ${B_1} = {B_2}$
Hence $\frac{{2M}}{{{x^3}}} = \frac{M}{{{y^3}}}{\rm{or}}\frac{{{x^3}}}{{{y^3}}} = 2\;{\rm{or}}\;\frac{x}{y} = {2^{1/3}}$
Standard 12
Physics
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