# A short bar magnet of magnetic movement   $5.25 \times 10 ^{-2} JT^{-1}$  is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45° with earth’s field on  its normal bisector.

Magnetic moment M is   $= 5.25\times 10^{-2}\ JT^{-1}$

Earth's field = 0.42 G.

The magnetic field due to magnet at equitorial line is given by :-

$B\ =\ \frac{\mu _0}{4\Pi }. \frac{M}{r^3}$

or                                                  $\frac{\mu _0}{4\Pi }. \frac{M}{r^3}\ =\ 0.42\times 10^{-4}$

or                                                    $r^3\ =\ 125\times 10^{-8}\ m$

$r\ =\ 0.05 m$

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