Join Community Community

Physics

n/a

If You Find Any Mistakes On Our Content or Want To make some Changes Then Report Now

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

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

 

Broadside on Position

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

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

Modefied By

Comments

#

Bug bounty – According to the online encyclopedia Wikipedia, the United States and India are the top countries from which researchers submit their bugs. India... Read Now

#

Bug bounty – According to the online encyclopedia Wikipedia, the United States and India are the top countries from which researchers submit their bugs. India... Read Now

Contents

We just added New Courses Notes Pdf

Search Pdf

We just added New Subjects

Search Subjects