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5.Magnetism and Matter
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A bar magnet of length $ 3 \,cm$ has points $A $ and $ B$ along its axis at distances of $24\, cm$ and $48\, cm$ on the opposite sides. Ratio of magnetic fields at these points will be
A
$8$
B
$1/2\sqrt 2 $
C
$3$
D
$4$
Solution
(a)Both points $ A$ and $B$ lying on the axis of the magnet and on axial position
$B \propto \frac{1}{{{d^3}}}$$⇒$ $\frac{{{B_A}}}{{{B_B}}} = {\left( {\frac{{{d_B}}}{{{d_A}}}} \right)^3} = {\left( {\frac{{48}}{{24}}} \right)^3} = \frac{8}{1}$
Standard 12
Physics
