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. 

Answers (1)

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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