Derive an expression for the magnetic field due to a bar magnet at an arbitrary point.

Derive an expression for the magnetic field due to a bar magnet at an arbitrary point.

SPREAD TO HELP OTHERS
  • 4
  •  
  • 1
  •  
  •  
  •  
  •  
  •  
    5
    Shares

In this article, we are going to derive an expression for the magnetic field due to a bar magnet at an arbitrary point around the bar magnet. So let’s get started…

Derivation for the magnetic field due to an bar magnet at an arbitrary point

Derive an expression for the magnetic field due to a bar magnet at an arbitrary point.
Fig.1, magnetic field due to a bar magnet at an arbitrary point, source: esaral.com

Let’s take a point P at a distance r from the center of the bar magnet. Join the line from point P to the center of the magnet and let this line make an angle \theta with the magnetic axis.

Magnetic field due to the bar magnet at an arbitrary point P has two magnetic field components i.e vertical components and horizontal components. Horizontal components of the magnetic field \vec{B_1} is because if you see the above figure then you will find that the point P is on the axial line of a virtual magnet S’N’ having magnetic moment 2M\cos\theta and the vertical components of the magnetic field \vec{B_2} is because if you again see the above figure then you will find that the point P is on the equatorial line of the virtual magnet S”N” with magnetic moment M\sin\theta.

Let \vec{B_1} is the horizontal and \vec{B_2} is the vertical components of the magnetic field. Then the expression of the magnetic field of both of the field components is:

    \[\vec{B_1}=\frac{\mu_0}{4\pi}\cdot\frac{2M\cos\theta}{r^3}\]

    \[\vec{B_2}=\frac{\mu_0}{4\pi}\cdot\frac{M\cos\theta}{r^3}\]

The total resultant magnetic field at point P is given as:

    \[\vec{B}=\sqrt{\vec{B_1}^2+\vec{B_2}^3}\]

Putting the values of \vec{B_1} and \vec{B_2} in above formula we get-

    \[\vec{B_1}=\frac{\mu}{4\pi}\cdot\frac{M}{r^3}\sqrt{4\cos^2\theta+\sin^2\theta}\]

    \[\vec{B_1}=\frac{\mu}{4\pi}\cdot\frac{M}{r^3}\sqrt{3\cos^2\theta+1}\]

From the above figure \tan\theta^{\prime} is given by \frac{B_2}{B_1} , so solving we get as below:

    \begin{equation*}\begin{split}\tan \theta^{\prime}&=\frac{B_{2}}{B_{1}}=\frac{\frac{\mu_{0}}{4 \pi} \frac{M \sin \theta}{r^{3}}}{\frac{\mu_{0}}{4 \pi} \frac{2 M \cos \theta}{r^{3}}}=\frac{1}{2} \tan \theta\\ \theta^{\prime}&=\tan ^{-1}\left(\frac{1}{2} \tan \theta\right)\end{split}\end{equation*}

Suggested reading:

Stay tuned with Laws Of Nature for more useful and interesting content.

Was this helpful

Leave a Reply