A short bar magnet moment `5.25 xx 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 in inclind at `45^(@)` with earth's field on (a) its normal bisector and (b) its axis . Magntiude of the earth's field at the place is given to be 0.42 G . Ignore the lenght of the magnet in comparison of the distance involved .

Explanation

To solve the problem, we need to find the distances from the center of a short bar magnet at which the resultant magnetic field is inclined at 45 degrees with respect to the Earth's magnetic field. We will consider two cases: (a) at the normal bisector of the magnet and (b) along its axis.

### Given Data:
- Magnetic moment of the magnet, \( m = 5.25 \times 10^{-2} \, \text{J/T} \)
- Magnitude of the Earth's magnetic field, \( B_E = 0.42 \, \text{G} = 0.42 \times 10^{-4} \, \text{T} \)

### (a) Distance from the center of the magnet at its normal bisector

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