COULOMB’S INVERSE SQUARE LAW OF MAGNETISM

Consider two bar magnets A and B as shown in Figure 3.11. When the north pole of magnet A and the north pole of magnet B or the south pole of magnet A and the south pole of magnet B are brought closer, they repel each other.

On the other hand, when the north pole of magnet A and the south pole of magnet B or the south pole of magnet A and the north pole of magnet B are brought closer, their poles attract each other.

This looks similar to Coulomb’s law for static charges studied in Unit I (opposite charges attract and like charges repel each other). So analogous to Coulomb’s law in electrostatics, we can state Coulomb’s law for magnetism (Figure 3.12) as follows:

Figure 3.11: Magnetic poles behave like electric charges – like poles repel and unlike poles attract

The force of attraction or repulsion between two magnetic poles is directly proportional to the product of their pole strengths and inversely proportional to the square of the distance between them.

Mathematically, we can write

where q_mA and q_mB are pole strengths of two poles and r is the distance between two magnetic poles.

In magnitude, \[ F = K{\frac{q_{m_{A}} q_{m_{B}}}{r^2}} \;\;\;\;\;\;\;\;(3.8) \]

where k is a proportionality constant whose valuedepends on the surrounding medium. In SI unit, the value of k for free space is k =µ₀/4π ~_~ 10^-7 H m^-1 , where μ_o is the absolute permeability of free space (air or vacuum) and H stands for henry.

Figure 3.12 Coulomb’s law – force between two magnetic poles

EXAMPLE 3.5

The repulsive force between two magnetic poles in air is 9 × 10^–3 N.

If the two poles are equal in strength and are separated by a distance of 10 cm, calculate the pole strength of each pole.

Solution:

The magnitude of the force between two poles is given by

Given : F = 9 × 10^–3N, r = 10 cm = 10 × 10^–2 m

Since q_mA = q_mB = q_m, we have

Consider a bar magnet NS as shown in Figure 3.13. Let N be the north pole and S be the south pole of the bar magnet, each of pole strength q_m and are separated by a distance of 2l. The magnetic field at a point C (lies along the axis of the magnet)

Figure 3.13 Magnetic field at a point along the axial line due to magnetic dipole

at a distance r from the geometrical centre O of the bar magnet can be computed by keeping unit north pole (q_mC = 1 A_m) at C.

The magnetic field at C due to the north pole is

where (r – l) is the distance between north pole of the bar magnet and unit north pole at C. The magnetic field at C due to the south pole is

where (r + l) is the distance between south pole of the bar magnet and unit north pole at C. The net magnetic field due to the magnetic dipole at point C

Since the magnitude of magnetic dipole moment is |p_m| = p_m = q_m.2l , the magnetic field at a point C can be written as

If the distance between two poles in a bar magnet is small (looks like short magnet) when compared to the distance between geometrical centre O of bar magnet and the location of point C (r »l),

Therefore, using equation (3.13) in equation (3.12), we get

where p_m = p_m i^{^}.

Consider a bar magnet NS as shown in Figure 3.14. Let N be the north pole and S be the south pole of the bar magnet, each with pole strength q_m and separated by a distance of 2l. The magnetic field at a point C (lies along the equatorial line) at a distance r from the geometrical centre O of the bar magnet can be computed by keeping unit north pole (q_mC = 1 A m) at C.

Figure 3.14 Magnetic field at a point along the equatorial line due to a magnetic dipole

The magnetic field at C due to the north pole is

where

Here,

The magnetic field at C due to the south pole is

Figure 3.15 Components of magnetic field

where, \[ \vec B_{S} = {\frac{\mu_\circ}{4\pi}} {\frac{q_{m}} {r'^{2}}} \]

From equations (3.15) and (3.16), the net magnetic field at point C due to the dipole is \[ \vec{B} = \vec B_{N} + \vec B_{S} \]

In a right angle triangle NOC as shown in Figure 3.14

Substituting equation (3.18) in equation (3.17), we get

Since, magnitude of magnetic dipole moment is |p_m| = p_m = q_m.2l and substituting in equation (3.19), the magnetic field at a point C is

If the distance between two poles in a bar magnet is small (looks like short magnet) when compared to the distance between geometrical centre O of bar magnet and the location of point C (r »l),

Therefore, using equation (3.21) in equation (3.20), we get

Since p_m i^{^}=p_m , the magnetic field at equatorial point is given by

Note that magnitude of B_axial is twice that of magnitude of B_equatorial and the direction of B_axial and B_equatorial are opposite.

EXAMPLE 3.6

A short bar magnet has a magnetic moment of 0.5 J T^–1.

Calculate magnitude and direction of the magnetic field produced by the bar magnet which is kept at a distance of 0.1 m from the centre of the bar magnet along (a) axial line of the bar magnet and (b) normal bisector of the bar magnet.

Solution

Given magnetic moment = 0.5 J T^–1 and distance r = 0.1 m

(a) When the point lies on the axial line of the bar magnet, the magnetic field for short magnet is given by

Hence, the magnitude of the magnetic field along axial is B_axial = 1 × 10^–4 T and direction is towards South to North.

(b) When the point lies on the normal bisector (equatorial) line of the bar magnet, the magnetic field for short magnet is given by

Hence, the magnitude of the magnetic field along axial is B_equatorial = 0.5 × 10^–4 T and direction is towards North to South. Note that magnitude of B_axial is twice that of magnitude of B_equatorial and the direction of B_axial and B_equatorial are opposite.