A short bar magnet has a magnetic moment of `0*48JT^-1`. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of `10cm` from the centre of the magnet on (i) the axis (ii) the equatorial line (normal bisector) of the magnet.

Explanation

To solve the problem, we need to find the magnetic field produced by a short bar magnet at a distance of 10 cm from its center on both the axis and the equatorial line. The magnetic moment (m) of the magnet is given as 0.48 J/T.

### Step 1: Magnetic Field on the Axis of the Magnet

1. **Formula for Magnetic Field on the Axis**: The magnetic field (B_A) at a point on the axis of a bar magnet is given by the formula:
\[ B_A = \frac{\mu_0}{4\pi} \cdot \frac{2m}{d^3} \]
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