The Neutral Points : Where Bar Magnet’s Field and Earth’s Magnetic Field Balance

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What are the Neutral Points in the Magnetic Field of a Bar Magnet?

How is the horizontal component of earth’s magnetic field related with the total intensity of earth’s magnetic field and the angle of dip?

In the discussions of the magnetic lines of force of a bar magnet, the influence of the earth’s magnetic field has not been taken into account so far. The lines of force for geomagnetic field at a place remain parallel to the magnetic meridian at that place and the direction of the lines of force is from south to north. The pattern of the lines of force near a bar magnet gets distorted under the influence of the geomagnetic field. A few cases are shown below.

N-pole pointing north: Let a bar magnet be placed along the magnetic meridian in such a manner that its N -pole points north [Fig.]. In this case, the pattern of the magnetic lines of force due to the combined effect of the geomagnetic field and the magnetic field due to the bar magnet is shown. From the

figure we see that, in the vicinity of magnet, the lines of force are curved. In this region, the influence of the bar magnet is more effective. Greater the distance from the magnet, lesser will be its influence but greater will be the influence of the geomagnetic field. At a sufficient distance from the magnet, its influence almost vanishes. In that region, the lines of force are due to geo-magnetic field only. Hence, those lines of force are straight, par-allel and directed from south to north.

At different points on the axis of the bar magnet, as the lines of force due to the bar magnet and geomagnetic field are in the same direction, the value of the magnetic intensity increases. But at different points on the perpendicular bisector of the mag-netic axis, as the lines of force due to geomagnetic field and the bar magnet are opposite in direction, the value of the magnetic intensity decreases.

On this perpendicular bisector there are two points X, X [Fig.] where the intensities due to geomagnetic field and the magnetic field of the magnet become equal and opposite. As a result, these two intensities cancel each other; i.e., at these two points, the resultant magnetic intensity becomes zero. These two points are known as neutral points. They lie at equal distance on either side of the magnet. A magnetic needle placed at any of these two neutral points, does not show any directive property. Naturally, no line of force passes through the neutral points.

Neutral point: A point in a magnetic field where the resultant magnetic intensity due to superposition of two or more magnetic fields becomes zero is called a neutral point.
When the N -pole of the bar magnet points north, at the neutral point,
F = H or, \(\frac{p_m}{\left(d^2+l^2\right)^{3 / 2}}\) = H [in CGS system]
[where, F = resultant intensity due to the bar magnet,
H = hori-zontal component of geomagnetic intensity,
p_m = magnetic moment of the bar magnet,
d = distance of the neutral point from the centre of the bar magnet and
2l = magnetic length of the bar magnet.]
For a tiny bar magnet, \(\frac{p_m}{d^3}\) = H. Usually if the value of d is more than ten times of l, then l² can be neglected compared to d².

S-pole pointing north: Let a bar magnet be placed along the magnetic meridian in such a way that its S -pole points north [Fig.], In this case, the pattern of the magnetic lines of force due to the combined effect of the geomagnetic field and magnetic field due to the bar magnet is shown in Fig.

Here the neutral points (X, X) lie on the axis of the bar magnet at equal distances from its two ends.
When the S -pole of the bar magnet points north, at neutral points,
F = H or, \(\frac{2 p_m d}{\left(d^2-l^2\right)^2}\) = H [in CGS system]
[where, F = intensity of the bar magnet,
H = horizontal component of the geomagnetic field,
p_m = magnetic moment of the
bar magnet, d = distance of the neutral point from the centre of the magnet and
2l = magnetic length of the bar magnet]
For a tiny bar magnet, \(\frac{2 p_m}{d^3}\) = H. If the value of d is more than
10 times of l, then l² can be neglected compared to d².

N-pole pointing east: If the north pole of a bar magnet points east, the pattern of magnetic lines of force due to the combined effect of geomagnetic field and the magnetic field of the bar magnet is shown in Fig. The two points X, X at the

north-west and south-east of the bar magnet are the neutral points. These are at equal distance from the centre of the magnet.

N-pole pointing west: If the north pole of a bar magnet points west, the pattern of the magnetic lines of force due to the combined effect of the geomagnetic field and magnetic field due to the bar magnet is shown in Fig. The two points X, X at the north-east and south-west of the magnet are the neutral points. Their distances from the centre of the magnet are equal.

Numerical Examples

Example 1.
The angle of dip at a place is 30° and the horizontal component of earth’s magnetic field at that place is 0.39 CGS units. Determine the vertical component of earth’s magnetic field at that place. (HS ‘04)
Solution:
Here, θ = 30° and H = 0.39 CGS units.
If the intensity of the earth’s magnetic field be I, then
H = Icos° or, I = \(\frac{H}{\cos \theta}\)
Vertical component,
V = Isinθ = \(\frac{H}{\cos \theta}\) ᐧ sinθ = Htanθ
= 0.39 × tan 30° = 0.39 × \(\frac{1}{\sqrt{3}}\) = 0.225 CGS units

Example 2.
At two places, the angles of dip are 30° N and 30° S and the intensity of the earth’s magnetic field is 0.42 Oe. Determine the horizontal and the vertical components of earth’s magnetic field at these two places and also Indicate their directions with the help of a diagram.
Solution:
At the first place, θ = 30° N [Fig.] and at the second place, θ = 30° S [Fig.].
Intensity of the earth’s magnetic field at both the places,
I = 0.42 Oe
∴ Horizontal component,
H = Icos30° = 0.42 × \(\frac{\sqrt{3}}{2}\) = 0.364 Oe
and vertical component,
V = Isin30° = 0.42 × \(\frac{1}{2}\) = 0.21 Oe

Example 3.
At a place, the horizontal and vertical components of earth’s magnetic field are 0.3 0e and 0.2 0e, respec tively. Determine the resultant Intensity and angle of dip there.
Solution:
Here, H = 0.3 Oe, V = 0.2 Oe.
If the resultant intensity is I then
I = \(\sqrt{H^2+V^2}\) = \(\sqrt{(0.3)^2+(0.2)^2}\) = \(\sqrt{0.13}\) = 0.3605 Oe
If the angle of dip is θ then,
tanθ = \(\frac{V}{H}\) = \(\frac{0.2}{0.3}\) = 0.6667
θ = tan^-1(0.6667) = 33.69°

Example 4.
The angle of dip and the horizontal component of earth’s magnetic field at a place are 30° S and 0.36 Oe. Determine the magnitude and direction of the vertical component of earth’s magnetic field at that place. (HS ‘02)
Solution:
Here, θ = 30° S. So, in equilibrium, the south pole of the magnetic needle leans downwards [Fig.].
We know that, tan θ = \(\frac{V}{H}\)

∴ V = vertical component of the earth’s magnetic field
= Htanθ
= 0.36tan30° [∵ H = 0.36 0e]
= 0.36 × \(\frac{1}{\sqrt{3}}\) = 0.208 Oe
Since the north pole of the magnetic needle lies above the horizontal line, the direction of the vertical component (V) will be vertically upwards.

Example 5.
At a place, the angle of declination is 30° E and the angle of dip is 45°N. Determine the horizontal and vertical components of the geomagnetic intensity in geographical meridian at that place. Given, the horizontal component of earth’s magnetic field at that place = 0.3 Oe. (HS ‘04)
Solution:
Here, δ = 30° E, θ = 45°N and H = 0.3 0e.
If the horizontal component of earth’s magnetic field in the geographical meridian be H’, according to Fig.

H’ = Hcosδ = 0.3 cos30° = 0.3 × \(\frac{\sqrt{3}}{2}\) = 0.2598 Oe
Vertical component remains the same in the geographical and the magnetic meridian planes, hence
V = Htanθ = 0.3 tan 45° = 0.3 × 1 = 0.3 Oe

Example 6.
The mass of a magnetic needle is 7.5 g and its magnetic moment is 98 units. To keep the magnetic needle horizontal in the northern hemisphere, what should be the position of its fulcrum with respect to its centre of gravtv? Vertical component of earth’s magnetic field
= 0.25 Oe.
Solution:
Let the fulcrum be kept at a distance x (towards the north pole) from the centre of gravity to keep the magnetic needle horizontal [Fig.].

If the length of the magnetic needle = 2l and the strength of each pole = m, magnetic moment, p_m = m ᐧ 2l.
∴ In equilibrium,
mV ᐧ 2l = W ᐧ x
or, V ᐧ p_m = W ᐧ x
or, x = \(\frac{V \cdot p_m}{W}\) = \(\frac{0.25 \times 98}{7.5 \times 980}\) = 0.0033 cm

Example 7.
The magnetic moment of a magnetic needle of mass 3.2 g is 980 CGS units. From which point should the needle be hung so that it will remain horizontal in the magnetic meridian? Horizontal component of the earth’s magnetic field at that place is 0.32 Oe and angle of dip = 45°N. [g = 980 cm ᐧ s^-2]
Solution:
Let the magnetic needle be hung from a point at a distance x from its centre of gravity (towards the north pole) [Fig.].

In equilibrium,
mV × 2l = W × x
or, V ᐧ p_m = W ᐧ x [∵ p_m = m ᐧ 2l]
or, x = \(\frac{V \cdot p_m}{W}\) = \(\frac{H \tan \theta \cdot p_m}{W}\) [∵ V = Htanθ]
= \(\frac{0.32 \times \tan 45^{\circ} \times 980}{3.2 \times 980}\) = 0.1 cm

Example 8.
Angle of dip at a place = θ; If the angle of dip in a vertical plane making angle δ with the magnetic meridian be θ’, show that, tanθ’: tanθ = secδ : 1.
Solution:
If the true dip angle in magnetic meridian be θ, the vertical and horizontal components of earth’s magnetic field be V and H, respectively then,
tanθ = \(\frac{V}{H}\)
In a vertical plane inclined at an angle δ with the magnetic meridian, horizontal component of intensity,
H’ = Hcosδ, apparent dip at that plane = θ’ and vertical component = V.
∴ tanθ’ = \(\frac{V}{H^{\prime}}\) = \(\frac{V}{H \cos \delta}\) = tanθ’ secδ
or, \(\frac{\tan \theta^{\prime}}{\tan \theta}\) = secδ
∴ tanθ’ – tanθ = secδ : 1

Example 9.
At a place, the apparent geomagnetic dip in a vertical plane is 40° and in another plane perpendicular to it is 30°. What is the real dip at the place?
Similar problem: If θ₁ is the angle of dip of the magnetic axis of a magnetic needle with horizontal at any vertical plane and θ₂ is that in another vertical plane at right angles to the former, prove that the real angle of dip, θ is given by cot²θ = cot²θ₁ + cot²θ₂.
Solution:
If the horizontal and vertical components of the earth’s magnetic field are H and V, respectively and the true dip angle at that place is θ then,
tanθ = \(\frac{V}{H}\) …… (1)

In Fig., M is the magnetic meridian and X, Y are two vertical planes inclined at right angle to each other. The angle between the planes X and M is δ. If the apparent dip in the plane X be θ₁ then,
tanθ₁ = \(\frac{V}{H \cos \delta}\) ……… (2)
If the apparent dip in the plane Y be θ₂ then,
tanθ₂ = \(\frac{V}{H \sin \delta}\) ……… (3)
From equations (2) and (3) we get,
cot²θ₁ + cot²θ₂ = \(\frac{H^2}{V^2}\) (cos²δ + sin²δ) = \(\frac{H^2}{V^2}\)
= cot²θ [from equation (1)]
Here, θ₁ = 40° and θ₂ = 30°.
∴ cot²θ = cot²40° + cot²30°
or, cot θ = \(\sqrt{1.42+3}\) = \(\sqrt{4.42}\) = 2.1
or, θ = cot^-1(2.1) = 25.46°

Example 10.
A bar magnet of length 6 cm is kept vertically with its north pale on the ground. If the distance of neutral point on the ground Is 8 cm from the north pole, what will be the magnetic moment of that magnet? [H = 0.36 CGS units] [HS ‘08]
Solution:
Let O he the neutral point [Fig.],
Here, NS = 6 cm, NO = 8 cm
So, SO = \(\sqrt{6^2+8^2}\) = 10 cm

If the pole-strength of the magnet NS is q_m, magnetic intensity at the point O
O due to the north pole,
H₁ = \(\frac{q_m}{O N^2}\), in the direction of OA [in CGS]
Again, due to the south pole magnetic intensity at the point O,
H₂ = \(\frac{q_m}{O S^2}\), in the direction of OS.
∴ Component of H₂ in the direction ON,
H₂cosθ = \(\frac{q_m}{O S^2} \cdot \frac{O N}{O S}\)
So, the horizontal magnetic intensity at the point O due to the entire magnet,
H₁ – H₂cosθ = \(\frac{q_m}{O N^2}\) – \(\frac{q_m}{O S^2} \cdot \frac{O N}{O S}\) = \(q_m\left(\frac{1}{8^2}-\frac{1}{10^2} \times \frac{8}{10}\right)\)
= q_m × 0.007625 Oe
Since, O is the neutral point, the magnetic intensity at that point due to the magnet will be equal but opposite to the horizontal component H of earth’s magnetic field.
∴ q_m × 0.007625 = 0.36 or, q_m = \(\frac{0.36}{0.007625}\) emu ᐧ cm
∴ Magnetic moment of the magnet
= q_m ᐧ NS = \(\frac{0.36}{0.007625}\) × 6 = 283.3 emu ᐧ cm²

Example 11.
A bar magnet of length 8 cm is placed on a horizontal plane in the magnetic meridian with its north pole pointing north. If the magnetic moment of the magnet be 90 CGS units and the horizontal component of earth’s magnetic field be 0.35 Oe, determine the positions of the neutral points. [HS ‘03]
Solution:
Let each of two neutral points he at a distance d from the centre of the magnet along its perpendicular bisector. If the magnetic moment of the magnet be p_m and its magnetic length be 2l, the magnetic intensity at the neutral points due to the magnet (in CGS system) is,
H_m = \(\frac{p_m}{\left(d^2+l^2\right)^{3 / 2}}\)
At the neutral points, the horizontal component H of earth’s magnetic field will be its equal and opposite.

Example 12.
At a place, the vertical component of earth’s magnetic field is \(\sqrt{3}\) times Its horizontal component. What will be the angle of dip at that place? [HS ‘06]
Solution:
If the intensity of earth’s magnetic field is I and the angle of dip is θ, then horizontal component of the earth’s magnetic field, H = Icosθ and vèrtical component, V = Isinθ.
According to the problem,
V = \(\sqrt{3}\)H or, \(\frac{V}{H}\) = \(\sqrt{3}\) or, \(\frac{I \sin \theta}{I \cos \theta}\) = \(\sqrt{3}\)
or, tanθ = \(\sqrt{3}\) = tan60° ∴ θ = 60°