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# A short bar magnet of magnetic moment $5.25 \times 10^{-2}\;J/T$ is placed with its axis perpendicular to the earth's field direction . At what distance from the center of the magnet , the resultant field is inclined at $45^{\circ}$ with earth's field at the place is given to be $0.42\;G$ Ignore the length of the magnet in companion to the distance involved

$\begin{array}{1 1} 6.3\;cm \\ 63\;cm \\ 0.63 \;cm \\ 6.3 \;m \end{array}$

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A)
Solution :
When the point lies on axial line .
Let the resultant magnetic field $B_r$ makes an angle of $45^{\circ}$
$\therefore B^1 = \large\frac{\mu_0}{4 \pi} \frac{2m}{r^3}$ (S to N)
$\therefore \tan 45^{\circ} = \large\frac{B^1 \sin 90^{\circ}}{B' \cos 90^{\circ}+Be}$
$1= \large\frac{B^1}{B^e}$
$=> B_e =B^1$
$0.42 \times 10^{-4} =\large\frac{\mu_0}{4 \pi } \times \frac{2m}{r^3}$
=> $0.42 \times 10^{-4}= \large\frac{10^{-7} \times 2 \times 5.25 \times 10^{-2}}{r^3}$
$r^3 =\large\frac{10^{-9} \times 2 \times 5.25 }{0.42 \times 10^{-4}}$
$\quad= 2.5 \times 10^{-5}$
$r= (2.5 \times 10^{-5})^{1/3}$
$r= 0.063\;m$ or $6.3 \;cm$
Answer : $6.3\;cm$