A short bar magnet of magnetic moment \( 5.25 \times 10^{-2}\,\mathrm{J/T} \) is placed with its axis perpendicular to earth's magnetic field direction. The distance of the point from the center of magnet on its perpendicular bisector, where the resultant field is \( 45^\circ \) to the earth's field will be: (Earth's field at that place is 0.42 G)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 5 \times 10^{-2}\,\mathrm{m} \)
On the perpendicular bisector (equatorial line) of a short magnet the field is parallel to the magnet's axis. Here the axis is perpendicular to the earth's field, so the magnet's field and the earth's field are at right angles. The resultant makes \( 45^\circ \) with the earth's field only when the two fields are equal in magnitude.
\( \displaystyle \frac{\mu_0}{4\pi}\,\frac{M}{r^{3}} = B_{E} \)
\( \mu_0/4\pi = 10^{-7}\,\mathrm{T\,m\,A^{-1}} \)
\( B_{E} = 0.42\,\mathrm{G} = 0.42\times10^{-4}\,\mathrm{T} \)
\( \displaystyle r^{3} = \frac{10^{-7} \times 5.25\times10^{-2}}{0.42\times10^{-4}} = 1.25\times10^{-4}\,\mathrm{m^{3}} = 125\times10^{-6}\,\mathrm{m^{3}} \)
\( \displaystyle r = 5\times10^{-2}\,\mathrm{m} \)