CBSE · Class 12 · Physics · Magnetism and MatterA short bar magnet has a magnetic moment of $0.48 \text{ J T}^{-1}$. Find the magnitude and direction of the magnetic field produced by the magnet at a distance of $10 \text{ cm}$ from the center of the magnet on (a) the axial line, and (b) the equatorial line (transverse line).

Given Data

• Magnetic dipole moment ($M$) = $0.48 \text{ J T}^{-1}$
• Distance ($r$) = $10 \text{ cm} = 0.1 \text{ m}$
• Permeability of free space ($\frac{\mu_0}{4\pi}$) = $10^{-7} \text{ T m A}^{-1}$

(a) Magnetic Field on the Axial Line ($B_{\text{axial}}$)

1. Formula: The magnetic field at an axial point for a short bar magnet is given by: $$B_{\text{axial}} = \frac{\mu_0}{4\pi} \frac{2M}{r^3}$$
2. Substitution: $$B_{\text{axial}} = 10^{-7} \times \frac{2 \times 0.48}{(0.1)^3}$$ $$B_{\text{axial}} = 10^{-7} \times \frac{0.96}{0.001}$$ $$B_{\text{axial}} = 10^{-7} \times 960$$ $$B_{\text{axial}} = 9.6 \times 10^{-5} \text{ T}$$
3. Direction: The direction of the magnetic field on the axial line is along the direction of the magnetic dipole moment vector (from South pole to North pole).

(b) Magnetic Field on the Equatorial Line ($B_{\text{equatorial}}$)

1. Formula: The magnetic field at an equatorial point for a short bar magnet is given by: $$B_{\text{equatorial}} = \frac{\mu_0}{4\pi} \frac{M}{r^3}$$
2. Substitution: $$B_{\text{equatorial}} = 10^{-7} \times \frac{0.48}{(0.1)^3}$$ $$B_{\text{equatorial}} = 10^{-7} \times \frac{0.48}{0.001}$$ $$B_{\text{equatorial}} = 10^{-7} \times 480$$ $$B_{\text{equatorial}} = 4.8 \times 10^{-5} \text{ T}$$
3. Direction: The direction of the magnetic field on the equatorial line is opposite to the direction of the magnetic dipole moment vector (from North pole to South pole).

Final Answer

• (a) Magnetic field on the axial line is $9.6 \times 10^{-5} \text{ T}$, directed along the axis from South to North.
• (b) Magnetic field on the equatorial line is $4.8 \times 10^{-5} \text{ T}$, directed parallel to the axis from North to South.