First slide

Magnetism

Question

Two magnets A and B are identical and these are arranged as shown in the figure. The (short magnets are considered) ratio of distance $d_{1}$ and $d_{2}$ will be,

Moderate

A

${(2 t a n Ѳ)}^{1 / 3}$
B

${(2 t a n Ѳ)}^{- 1 / 3}$
C

${(2 c o t Ѳ)}^{- 1 / 3}$
D

${(2 c o t Ѳ)}^{1 / 3}$

Solution

$T h e m a g n e t i c n e e d l e w i l l a l i g n a l o n g t h e t o t a l m a g n e t i c f i e l d . I n E q u i l i b r i u m, t a n θ = \frac{B_{a x i a l}}{B_{e q u a t o r i a l}} = \frac{B_{1}}{B_{2}}$
$\Rightarrow B_{1} = B_{2} t a n θ \Rightarrow \frac{μ_{o}}{4 π} \frac{2 M}{{d_{1}}^{3}} = \frac{μ_{o}}{4 π} \frac{M}{{d_{2}}^{3}} t a n θ$
$\Rightarrow \frac{d_{1}}{d_{2}} = {[\frac{2}{t a n θ}]}^{\frac{1}{3}} = {[2 c o t θ]}^{\frac{1}{3}}$

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