A short bar magnet of mangetic moment `5*25xx10^-2JT^-1` is placed with its axis perpendicular to earth's field direction. At what distance from the centre of the magnet, is the resultant field inclined at `45^@` with earth's field on (i) its normal bisector, (ii) its aixs? Magnitude of earth's field at the place `0*42G`. Ignore the length of the magnet in comparison to the distances involved.

Explanation

To solve the problem, we need to calculate the distance from the center of the magnet where the resultant magnetic field is inclined at \( 45^\circ \) with respect to the Earth's magnetic field.

Given:
- Magnetic moment \( M = 5 \times 25 \times 10^{-2} \, \text{JT}^{-1} = 1.25 \times 10^{-1} \, \text{JT}^{-1} \)
- Earth's magnetic field \( B_E = 0.42 \, \text{G} = 0.42 \times 10^{-4} \, \text{T} \) (since \( 1 \, \text{G} = 10^{-4} \, \text{T} \))
- The angle of inclination \( \theta = 45^\circ \)

### Step 1: Calculate the magnetic field due to the bar magnet
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