Question
Derive an expression for the magnetic field due to a bar magnet at an arbitrary point.
✓
Answer
i. Consider a bar magnet of magnetic moment $\vec{m}$ with centre at $O$ as shown in figure and let $P$ be any point in its magnetic field.
ii. Magnetic moment $\vec{m}$ is resolved into components along $\vec{r}$ and perpendicular to $\vec{r}$.
iii. For the component $m \cos \theta$ along $\vec{r}$, the point P is an axial point.
iv. For the component $m \sin \theta$ perpendicular to $\vec{r}$, the point P is an equatorial point at the same distance $\vec{r}$.
v. For a point on the axis, $B _{ a }=\frac{\mu_0}{4 \pi} \frac{2 m}{ r ^3}$
Here
$B _{ a }=\frac{\mu_0}{4 \pi} \frac{2 m \cos \theta}{r^3}$
directed along $m \cos \theta$.
vi. For point on equator,
$B _{ a }=\frac{\mu_o}{4 \pi} \frac{m \sin \theta}{r^3}$
directed opposite to $m \sin \theta$
vii. Thus, the magnitude of the resultant magnetic field $B$, at point $P$ is given by
$\begin{aligned}
& B & =\sqrt{B_a^2+B_{e q}^2} \\
\therefore & B & =\sqrt{\left(\frac{\mu_0}{4 \pi} \frac{2 m \cos \theta}{r^3}\right)^2+\left(\frac{\mu_0}{4 \pi} \frac{ m \sin \theta}{r^3}\right)^2} \\
\therefore & B & =\frac{\mu_o}{4 \pi} \frac{ m }{ r ^3} \sqrt{(2 \cos \theta)^2+(\sin \theta)^2} \\
\therefore & B & =\frac{\mu_o}{4 \pi} \frac{ m }{ r ^3} \sqrt{3 \cos ^2 \theta+1}
\end{aligned}$
viii. Let a be the angle made by the direction of $\vec{B}$ with $\vec{r}$. Then, by using equation (1) and equation (2),
$\tan \alpha=\frac{B_{\text {eq }}}{B_a}=\frac{1}{2}(\tan \theta)$
The angle between directions of $\vec{B}$ and $\vec{m}$ is then $(\theta+ a )$.
ii. Magnetic moment $\vec{m}$ is resolved into components along $\vec{r}$ and perpendicular to $\vec{r}$.
iii. For the component $m \cos \theta$ along $\vec{r}$, the point P is an axial point.
iv. For the component $m \sin \theta$ perpendicular to $\vec{r}$, the point P is an equatorial point at the same distance $\vec{r}$.
v. For a point on the axis, $B _{ a }=\frac{\mu_0}{4 \pi} \frac{2 m}{ r ^3}$
Here
$B _{ a }=\frac{\mu_0}{4 \pi} \frac{2 m \cos \theta}{r^3}$
directed along $m \cos \theta$.
vi. For point on equator,
$B _{ a }=\frac{\mu_o}{4 \pi} \frac{m \sin \theta}{r^3}$
directed opposite to $m \sin \theta$
vii. Thus, the magnitude of the resultant magnetic field $B$, at point $P$ is given by
$\begin{aligned}
& B & =\sqrt{B_a^2+B_{e q}^2} \\
\therefore & B & =\sqrt{\left(\frac{\mu_0}{4 \pi} \frac{2 m \cos \theta}{r^3}\right)^2+\left(\frac{\mu_0}{4 \pi} \frac{ m \sin \theta}{r^3}\right)^2} \\
\therefore & B & =\frac{\mu_o}{4 \pi} \frac{ m }{ r ^3} \sqrt{(2 \cos \theta)^2+(\sin \theta)^2} \\
\therefore & B & =\frac{\mu_o}{4 \pi} \frac{ m }{ r ^3} \sqrt{3 \cos ^2 \theta+1}
\end{aligned}$
viii. Let a be the angle made by the direction of $\vec{B}$ with $\vec{r}$. Then, by using equation (1) and equation (2),
$\tan \alpha=\frac{B_{\text {eq }}}{B_a}=\frac{1}{2}(\tan \theta)$
The angle between directions of $\vec{B}$ and $\vec{m}$ is then $(\theta+ a )$.
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