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.
Text Solution
AI Generated Solution
To solve the problem, we will calculate the magnetic field produced by a short bar magnet at a distance of 10 cm (0.1 m) from its center on both the axis and the equatorial line.
### Given:
- Magnetic moment (M) = 0.48 J T⁻¹
- Distance (d) = 10 cm = 0.1 m
- Permeability of free space (μ₀) = 4π × 10⁻⁷ T m/A (constant)
### (i) Magnetic Field on the Axis of the Magnet:
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