Derive an expression of the magnetic field at the axial point of a bar magnet class 12

In this article, we will derive an expression for the magnetic field at a point on the axis of a bar magnet, so let’s get started…

Derivation of the magnetic field at the axial point of a bar magnet class 12

[latexpage]Let SN be a bar magnet of length $2l$ and magnetic pole strength of $q_m$. Suppose the point at which we wish to find the magnetic field, lies on the axis of the magnet at the distance $r$ from the center of the bar magnet. Here, P is the point where we want to find the magnetic field. See figure below:

Let $B_1$ be the magnetic field experienced by the point P due to the North pole of the bar magnet and $B_2$ be the magnetic field experienced by the point P due to the South pole of the bar magnet. The net magnetic field $(B)$ experienced by the point P (lie on the axis of a bar magnet) is the algebraic sum of all the magnetic field acted on it, i.e $$B_{axial}=B_1 + B_2$$

So magnetic field due to the North pole of the bar magnet at the point P is: $$B_1=\frac{\mu_0}{4\pi}\cdot\frac{q_m}{(r-l)^2}\quad\text{along}\:\overrightarrow{NP}$$ And the magnetic field due to South pole of the bar magnet at the point P is given as: $$B_2=\frac{\mu_0}{4\pi}\cdot\frac{q_m}{(r+l)^2}\quad\text{along}\:\overrightarrow {PS}$$

So the net magnetic field at the point P is given as: $$B_{axial}=B_1 + (-B_2)$$

Substitute the value of $B_1$ and $B_2$ in the above sum, we get-

$$B_{axial}=\left[\frac{\mu_0}{4\pi}\cdot\frac{q_m}{(r-l)^2}-\frac{\mu_0}{4\pi}\cdot\frac{q_m}{(r+l)^2}\right]$$ $$B_{axial}=\frac{\mu_0q_m}{4\pi}\left[\frac{1}{(r-l)^2}-\frac{1}{(r+l)^2}\right]$$ $$B_{axial}=\frac{\mu_0q_m}{4\pi}\left[\frac{4rl}{(r^2-l^2)^2}\right]$$

We know that $q_m\cdot 2l=M$, where M is the magnetic dipole moment of this magnetic dipole (bar magnet), so above equation can be rewritten as: $$B_{axial}=\frac{\mu_0}{4\pi}\cdot\frac{2Mr}{(r^2-l^2)^2}$$

For a short bar magnet, for which $(l

It is clearly visible that the direction of the magnetic field at the axial point due to a bar magnet is same as the direction of the magnetic dipole moment, i.e from S-pole to N-pole, so we can write: $$B_{axial}=\frac{\mu_0}{4\pi}\cdot\frac{2 \overrightarrow{M}}{r^3}$$

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