NCERT · Class 12 · Physics · Magnetism and MatterDefine the intensity of magnetic field (magnetic field strength) on the axial line of a bar magnet. Derive an expression for the magnetic field at a point situated at a distance $d$ along its axial line. Also, calculate the magnetic field intensity at a distance of 0.1 m from the center of a short bar magnet having a magnetic dipole moment of $0.4 \, \text{A}\cdot\text{m}^2$ along its axial line.

Part 1: Theory and Derivation

Definition of Magnetic Field Intensity on Axial Line:\nThe magnetic field intensity at a point on the axial line of a bar magnet is defined as the magnetic force experienced by a unit north pole placed at that point.

Derivation for Axial Magnetic Field:\nLet us consider a bar magnet of length $2l$ and pole strength $m$. The magnetic dipole moment of the bar magnet is given by $M = m \times 2l$. Let $P$ be a point on the axial line of the magnet at a distance $d$ from its center $O$.

• Magnetic field at point $P$ due to the North pole ($N$):\nThe distance of point $P$ from the North pole is $(d - l)$. $$B_1 = \frac{\mu_0}{4\pi} \frac{m}{(d - l)^2} \quad (\text{directed away from } N)$$

• Magnetic field at point $P$ due to the South pole ($S$):\nThe distance of point $P$ from the South pole is $(d + l)$. $$B_2 = \frac{\mu_0}{4\pi} \frac{m}{(d + l)^2} \quad (\text{directed towards } S)$$

• Net Magnetic Field ($B$):\nSince $B_1$ and $B_2$ act along the same straight line in opposite directions, the magnitude of the resultant magnetic field is: $$B = B_1 - B_2$$ $$B = \frac{\mu_0 m}{4\pi} \left[ \frac{1}{(d - l)^2} - \frac{1}{(d + l)^2} \right]$$ $$B = \frac{\mu_0 m}{4\pi} \left[ \frac{(d + l)^2 - (d - l)^2}{(d^2 - l^2)^2} \right] = \frac{\mu_0 m}{4\pi} \left[ \frac{4dl}{(d^2 - l^2)^2} \right]$$ \nSince $M = m \times (2l)$, we can rewrite this as: $$B = \frac{\mu_0}{4\pi} \frac{2Md}{(d^2 - l^2)^2}$$ \nFor a short bar magnet where $l \varpropto d$ (i.e., $l \ll d$), $l^2$ can be neglected in comparison to $d^2$: $$B = \frac{\mu_0}{4\pi} \frac{2M}{d^3}$|

Part 2: Numerical Problem

Given data:

• Magnetic dipole moment ($M$) = $0.4 , \text{A}\cdot\text{m}^2$
• Distance ($d$) = $0.1 , \text{m}$
• Permeability of free space ($\frac{\mu_0}{4\pi}$) = $10^{-7} , \text{T}\cdot\text{m/A}$

Formula:\nFor a short bar magnet on the axial line: $$B = \frac{\mu_0}{4\pi} \frac{2M}{d^3}$$

Step-by-Step Calculation:

1. Substitute the given values into the formula: $$B = 10^{-7} \times \frac{2 \times 0.4}{(0.1)^3}$$
2. Simplify the numerator: $$2 \times 0.4 = 0.8$$
3. Simplify the denominator: $$(0.1)^3 = (10^{-1})^3 = 10^{-3}$$
4. Divide the terms: $$B = 10^{-7} \times \frac{0.8}{10^{-3}}$$ $$B = 10^{-7} \times 0.8 \times 10^3$$ $$B = 0.8 \times 10^{-4} , \text{T}$$ $$B = 8 \times 10^{-5} , \text{Tesla (T)}$$

Answer:\nThe magnetic field intensity at the given point is $8 \times 10^{-5} , \text{T}$.