A short magnet of moment 6.75, Am^2 produces a neutral point on its axis. If horizontal component of

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5.Magnetism and Matter

medium

A short magnet of moment $6.75\, Am^2$ produces a neutral point on its axis. If horizontal component of earth's magnetic field is $5 \times {10^{ - 5}}\,Wb/{m^2}$, then the distance of the neutral point should be.....$cm$

$10$

$20$

$30$

$40$

Solution

(c)At neutral point
$\left| \begin{array}{c}\,{\rm{Magnetic field due }}\\{\rm{to magnet}}\end{array} \right| = \,\left| \begin{array}{c}{\rm{Magnetic field due }}\\{\rm{to earth}}\end{array} \right|$$\frac{{{\mu _0}}}{{4\pi }}.\frac{{2M}}{{{d^3}}} = 5 \times {10^{ – 5}} \Rightarrow {10^{ – 7}} \times \frac{{2 \times 6.75}}{{{d^3}}} = 5 \times {10^{ – 5}}$
$ \Rightarrow d = 0.3\;m = 30\;cm$

Standard 12

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