Question:

Find the pole strength of a magnet of length 2 cm, if the magnetic field strength \( B \) at distance 10 cm from the centre of a magnet on the axial line of the magnet is \( 10^{-4} \, \text{T} \).

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For a magnet, the magnetic field strength on the axial line can be calculated using the formula \( B = \frac{{\mu_0 \, m}}{{4 \pi r^3}} \), where \( m \) is the pole strength of the magnet and \( r \) is the distance from the centre of the magnet.
Updated On: May 8, 2025
  • 25 Am
  • 100 Am
  • \( 5 \times 10^{-2} \, \text{Am} \)
  • \( 1 \times 10^{-4} \, \text{Am} \)
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The Correct Option is A

Solution and Explanation

For a bar magnet, the magnetic field on the axial line at a distance \( r \) from the centre is given by the formula: \[ B = \frac{{\mu_0 \, m}}{{4 \pi \, r^3}} \left( 2 \right) \] where: - \( B \) is the magnetic field strength, - \( \mu_0 \) is the permeability of free space (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \)), - \( m \) is the magnetic pole strength, - \( r \) is the distance from the center of the magnet to the point where the field strength is measure(D) Given: - \( B = 10^{-4} \, \text{T} \), - \( r = 0.1 \, \text{m} \), - The length of the magnet is 2 cm, so the pole strength \( m \) is to be determine(D) Substitute the given values into the formula: \[ 10^{-4} = \frac{{(4 \pi \times 10^{-7}) \, m}}{{4 \pi \, (0.1)^3 \times 2}} \] Simplifying the equation: \[ 10^{-4} = \frac{{10^{-7} \, m}}{{8 \times 10^{-3}}} \] Solving for \( m \): \[ m = 10^{-4} \times 8 \times 10^{-3} = 25 \, \text{Am} \] Thus, the pole strength of the magnet is \( 25 \, \text{Am} \).
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