Question

The magnitude of the axial field due to a short bar magnet at a distance of 50 cm from its mid-point is

• A. \( 3.2 \times 10^{-7} \, \text{T} \)
• B. \( 1.6 \times 10^{-7} \, \text{T} \)
• C. \( 6.4 \times 10^{-7} \, \text{T} \)
• D. \( 4.8 \times 10^{-7} \, \text{T} \)

Hint:

The magnetic field at a point along the axis of a short bar magnet decreases with the cube of the distance from the center.

Answer 1

The Correct Option is C

The magnetic field due to a short bar magnet at a distance \( r \) from the center is given by the formula: \[ B = \frac{\mu_0 m}{2 \pi r^3} \] Where: - \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) is the permeability of free space, - \( m = 0.4 \, \text{Am}^2 \) is the magnetic moment of the bar magnet, - \( r = 0.5 \, \text{m} \) is the distance from the center. Substituting the values: \[ B = \frac{4\pi \times 10^{-7} \times 0.4}{2 \pi (0.5)^3} = 6.4 \times 10^{-7} \, \text{T} \] Thus, the magnetic field is \( \boxed{6.4 \times 10^{-7} \, \text{T}} \).