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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Magnetism and Matter
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Suppose a magnet is suspended by a vertical string, attached to its middle point. Given that the horizontal component of earth's magnetic field is $25\;\mu T$ and the vertical component of of earth's magnetic field is $40\;\mu T$, find the position in which the magnet can stay in equilibrium.

$\begin{array}{1 1} (A) \theta = \sin^{-1} \large (\frac{8}{5}) \\(B) \theta = \cos^{-1} \large (\frac{8}{5}) \\(C) \theta = \tan^{-1} \large (\frac{8}{5}) \\ (D) \theta = \tan^{-1} \large (\frac{5}{8}) \\ \end{array}$

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The vertical component of the earth's magnetic field is given by $B_v = B \sin \theta I$ and the horizontal component by $B_h = B \cos \theta I$
$\Rightarrow \large\frac{B_v}{B_h} = \large\frac{B \sin \theta I}{B \cos \theta I} $$ \rightarrow \tan \theta =\large\frac{B_v}{B_h}$
$\Rightarrow \tan \theta = \large\frac{40}{25} $$ =\large\frac{8}{5}$$ \rightarrow \theta = tan^{-1} \large (\frac{8}{5})$ $ \rightarrow \theta = 58 ^{\circ}$
This is the angle that the magnetic field makes with Earth's magnetic meredian. For any magnet to be stable in equilibrium, it needs to stay in this direction, thus making this angle with the horizontal plane.
answered Mar 14, 2014 by balaji.thirumalai
 

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