 ## Magnetic Field due to a Bar Magnet

Subject: Physics

#### Overview

The measurements at two points are important: at a point on the axis and at a point on the equatorial line of the bar magnet. These are called end-on position and broadside-on position. This note provides us an information on magnetic field due to a bar magnet.

### Magnetic field due to bar magnet

The strength of magnetic field at any point around a bar magnet can be calculated. However, the measurements at two points are important: at a point on the axis and at a point on the equatorial line of the bar magnet. These are called end-on position and broadside-on position respectively.

#### End-on position

Consider a bar magnet of length 2l and pole strength m. Suppose a point P on the axis of the magnet at a distance d from its center. (d –l) is the distance of P form the N-pole of the magnet. The magnetic field intensity at P due to the north-pole of the magnet is

\begin{align*} B_1 &= \frac {\mu _o}{4\pi }\: \frac {m}{r^2} \\ &= \frac {\mu _o}{4\pi }\: \frac {m}{(d – l)^2} \\\end{align*}

which is directly away from N-pole. Since the south of the magnet is at a distance r = d + l from P, so magnetic field intensity at P due to S-pole is

\begin{align*} B_2 &= \frac {\mu _o}{4\pi }\: \frac {m}{r^2} \\ &= \frac {\mu _o}{4\pi }\: \frac {m}{(d + l)^2} \\\end{align*} which is direct towards, the S-pole of the magnet. The magnetic field intensity B at P is the resultant of these two fields,

\begin{align*} B &= B_1 + (-B_2) \\ &= B_1 – B_2 \\ &= \frac {\mu _o m}{4\pi } \left [\frac {1}{(d – l)^2} - \frac {1}{d + l^2} \right ] \\ &=\frac {\mu _o m}{4pi } \left [ \frac {4ld}{(d^2 – l^2)^2}\right ] \\ &= \frac {\mu _o 2Md}{4\pi (d^2 – l^2)^2} \\ \end{align*}

where $M = m \times 2l$, the magnetic moment of the bar magnet. So, the magnetic field at a point on the axis of a bar magnet is

$$B = \frac {\mu _o Md}{2\pi (d^2 – l^2 )^2}$$

If the length of the magnet is very small, d>>I and the magnetic field intensity is

$$B = \frac {\mu _o M}{2\pi d^3}$$

Suppose a point P is on the equatorial line of the bar magnet. The equatorial line of the magnet is the line perpendicular to the axis of the magnet which bisects the magnet. Let d be the distance of the point P from the centre of the magnet and P due to the North Pole is

\begin{align*} B_1 &= \frac {\mu _o}{4\pi }\: \frac {m}{r^2} \\ &= \frac {\mu _o}{4\pi }\: \frac {m}{(d^2 + l^2)} \\\end{align*}

directed away from N-pole. the magnetic field B2 at P due to S-pole is

\begin{align*} B_2 &= \frac {\mu _o}{4\pi }\: \frac {m}{r^2} \\ &= \frac {\mu _o}{4\pi }\: \frac {m}{(d^2 + l^2)} \\\end{align*}

directed towards S-pole. These fields have different directions, but the same magnitude as shown in the figure. Let $\angle PSO = \theta$ and by symmetry, $\angle PNO = \theta$. The angle between B1 and B2 is then $2\theta$. The resultant magnetic field, B at P is given by

\begin{align*} B^2 &= B_1^2 + B_2^2 + 2B_1B_2 \cos \: 2\theta \\ \text {Since,} \: B_1 = B_2 \: \text {in magnitude, so} \\ B^2 &= B_1^2 (2 + 2\cos 2\theta )&= 2B_1^2(1 + \cos 2\theta ) \\ &= 4B_1^2 \cos ^2 \theta \\ \end{align*}

\begin{align*} \text {we have} \:\cos \theta &= \frac {1}{\sqrt {d^2 + l^2}} \\ \text {So,} \\ B &= 2B_1\cos \theta \\ &= 2 \frac {\mu _o}{4\pi } \frac {m}{(d^2 + l^2)}\frac {1}{\sqrt {d^2 + l^2}} \\ &=\frac {\mu _o }{4\pi } \frac {2ml}{(d^2 + l^2 )^{3/2}} \\ &=\frac {\mu _o }{4\pi } \frac {M}{d^2 + l^2 )^{3/2}} \\\end{align*}

The direction of B is parallel to the axis of the magnet, from north to south pole. If the magnet is very short, $d >> l$, and the magnetic field at P is

$$B = \frac {\mu _o}{4\pi } \frac {M}{d^3}$$

Reference

Manu Kumar Khatry, Manoj Kumar Thapa, et al.Principle of Physics. Kathmandu: Ayam publication PVT LTD, 2010.

S.K. Gautam, J.M. Pradhan. A text Book of Physics. Kathmandu: Surya Publication, 2003.

##### Things to remember

The measurements at two points are important: at a point on the axis and at a point on the equatorial line of the bar magnet. These are called end-on position and broadside-on position.

So, the magnetic field at a point on the axis of a bar magnet is $B = \frac {\mu _o Md}{2\pi (d^2 – l^2 )^2}$

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