A bar magnet `30cm` long is placed in magnetic meridian with its north pole pointing south. The neutral point is observed at a distance of `30cm` from its centre. Calculate the pole strength of the magnet. Given horizontal component of earth's field is `0*34G`.
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To solve the problem step by step, we will follow the logical progression of calculations based on the information provided.
### Step 1: Understand the setup
We have a bar magnet that is 30 cm long, placed in the magnetic meridian with its north pole pointing south. The neutral point is located 30 cm from the center of the magnet. The horizontal component of the Earth's magnetic field (H) is given as 0.34 G.
### Step 2: Convert units
Convert the given values into SI units:
- Length of the magnet (2L) = 30 cm = 0.30 m
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