A magnet the size of the Earth
A compass needle points north everywhere in India, from Kanyakumari to Leh. So the Earth has a magnetic field over its whole surface. Near the surface the field is weak, between about \(25\) and \(65\,\mu\text{T}\). That is about a hundredth of the field at the face of a fridge magnet. Its shape is close to the field of a bar magnet placed at the centre of the Earth.
That imagined magnet does not lie along the spin axis. It is tilted by about \(10°\), so the magnetic poles are not the geographic poles. The north end of a compass needle points to the magnetic pole in the northern hemisphere. Unlike poles attract, so that magnetic pole is a south pole of the Earth's magnet. The name "north magnetic pole" means the pole near the geographic north, not its polarity.
The magnet is imagined. The core of the Earth is molten iron at several thousand kelvin. Lesson 5 shows that iron cannot stay magnetised at such a temperature. The field is made by electric currents that flow in the moving liquid iron of the outer core. Because those currents change slowly, the magnetic poles wander, in recent decades by tens of kilometres a year. Every few hundred thousand years the field even reverses; the last reversal was about 780 000 years ago.
The north end of a compass needle points towards the magnetic pole in the Arctic. What kind of pole of the Earth's magnet is that?
Three numbers for a field
A field at a place is a vector in three dimensions. It takes three numbers to fix it. For the Earth's field the three elements of the Earth's field are chosen so that each can be read with a needle.
The vertical plane that holds a freely pivoted compass needle is the magnetic meridian of the place. The vertical plane through the geographic north-south line is the geographic meridian. The angle between the two is the declination, \(D\). A declination of \(10°\,\text{E}\) means the needle settles \(10°\) east of geographic north. Across India the declination is small, within about two degrees of zero.
In the magnetic meridian, the field itself points below the horizontal in the northern hemisphere. The angle it makes with the horizontal is the dip, \(\delta\), also called the inclination. The field splits into a horizontal part \(B_H\) along the meridian and a vertical part \(B_V\). A needle pivoted on a horizontal axis, free to turn in the vertical plane of the meridian, points along \(\vec B\) itself. This is a dip circle, and it reads \(\delta\) directly.
At the magnetic equator the needle lies flat and all the field is horizontal. Move towards the pole and the needle tilts down, \(B_H\) shrinks and \(B_V\) grows. At the pole the needle stands vertical and \(B_H\) is zero. The Delhi chip sets the needle at \(42°\), with \(B_H = 3.6\) and \(B_V = 3.24\) in units of \(10^{-5}\,\text{T}\). The figure's perfect dipole places Delhi at a magnetic latitude of \(24°\). The real field is less tidy, so the figure's latitudes are a model.
(a) At Delhi \(B_H = 3.6 \times 10^{-5}\,\text{T}\) and the dip is \(42°\). Find \(B_V\) and the total field \(B\). (b) At another place \(B_H = 4 \times 10^{-5}\,\text{T}\) and the dip is \(37°\); take \(\tan 37° = 0.75\). Find \(B_V\) and \(B\).
(a) \(B_V\) \(= B_H\tan\delta\) \(= 3.6 \times 10^{-5} \times 0.900\) \(= 3.24 \times 10^{-5}\,\text{T}\). \(B\) \(= B_H/\cos\delta\) \(= 3.6 \times 10^{-5}/0.743\) \(= 4.84 \times 10^{-5}\,\text{T}\).
(b) \(B_V = 4 \times 10^{-5} \times 0.75 = 3 \times 10^{-5}\,\text{T}\). \(B = \sqrt{4^2 + 3^2} \times 10^{-5} = 5 \times 10^{-5}\,\text{T}\). Check: \(B\cos 37° = 5 \times 0.8 = 4\), as given.
Values used: Delhi \(B_H = 3.6 \times 10^{-5}\,\text{T}\) at a dip of \(42°\); part (b) takes the assignment sheet's \(4 \times 10^{-5}\,\text{T}\) at \(37°\).
A compass needle turns only about a vertical pivot. The vertical part of the field only tries to tip the needle, and the pivot holds it level. So a compass feels only \(B_H\), and it settles along the magnetic meridian. Where the declination is \(10°\,\text{E}\), the compass points \(10°\) east of true north. A navigator adds the declination to a compass reading to find true north.
At a place the Earth's total field is \(5 \times 10^{-5}\,\text{T}\) and the dip is \(30°\). Find the horizontal component, in units of \(10^{-5}\,\text{T}\).
Which part of the Earth's field sets the direction of an ordinary compass needle?
Apparent dip
A dip circle reads the true dip only when its plane is the magnetic meridian. Turn it so that its plane makes an angle \(\theta\) with the meridian. The needle can move only in that plane, so it responds only to the parts of \(\vec B\) that lie in the plane. The vertical part \(B_V\) lies in every vertical plane. Of \(B_H\), only the part \(B_H\cos\theta\) lies in the new plane.
\[ B_{\parallel} = B_H\cos\theta \]
Why. the horizontal field makes the angle \(\theta\) with the plane of the circle; its part along the plane is \(B_H\cos\theta\).
\[ \tan\delta' = \dfrac{B_V}{B_H\cos\theta} \]
Why. the needle sets along the resultant of the two parts in its plane, \(B_V\) down and \(B_H\cos\theta\) along.
\[ \tan\delta' = \dfrac{\tan\delta}{\cos\theta} \]
Read it. \(B_V/B_H = \tan\delta\), the true dip.
Since \(\cos\theta\) is at most 1, the apparent dip is never less than the true dip. At \(\theta = 60°\), \(\tan\delta' = 0.75/0.5 = 1.5\), so \(\delta' = 56.3°\). At \(\theta = 90°\) the plane is east-west and no part of \(B_H\) lies in it. The needle then stands vertical, whatever the true dip. This is how the magnetic meridian is found: turn the circle until the needle is vertical, then turn it through a further \(90°\).
A dip circle reads an apparent dip of \(45°\) in one vertical plane and \(60°\) in the vertical plane at right angles to it. Find the true dip.
Let the first plane make \(\theta\) with the meridian; the second makes \(90° - \theta\). From \(\cot\delta' = \cot\delta\cos\theta\): \(\cot\delta_1 = \cot\delta\cos\theta\) and \(\cot\delta_2 = \cot\delta\sin\theta\).
Square and add: \(\cot^2\delta_1 + \cot^2\delta_2 = \cot^2\delta\). So \(\cot^2\delta\) \(= \cot^2 45° + \cot^2 60°\) \(= 1 + \tfrac13\) \(= \tfrac43\).
\(\cot\delta = 1.155\), so \(\tan\delta = 0.866\) and \(\delta = 40.9°\). The true dip is less than both apparent dips, as it must be.
At a place the true dip is \(45°\). A dip circle is set with its plane at \(30°\) to the magnetic meridian. Find the apparent dip, in degrees.
The magnetic equator and the poles
Carry a dip circle from the far south of India to Ladakh. Near Thiruvananthapuram the needle lies almost flat, and the dip is close to zero. The line of zero dip round the globe is the magnetic equator. Moving north, the needle tilts down more and more. At the magnetic pole, now in the Arctic Ocean, the field is vertical: the dip is \(90°\) and the needle stands straight down.
At a magnetic pole \(B_H = 0\), so an ordinary compass has nothing to turn it and points anywhere. On the magnetic equator \(B_V = 0\), and all the field is horizontal. Fig 2.1 shows both ends of this change round one globe.
For a perfect dipole Earth, the dip at magnetic latitude \(\lambda\) obeys \(\tan\delta = 2\tan\lambda\). It follows from the dipole field in the deeper note of Lesson 1, with the moment pointing south. The dip needle figure above uses this rule. The real Earth has extra, smaller parts in its field, so measured dips differ from it by several degrees.
At a certain place a dip needle, set in the magnetic meridian, stands exactly vertical. What is true there?
A compass needle has a magnetic moment of about \(0.01\,\text{A m}^2\). Estimate the largest torque the Earth's field can exert on it in Delhi, in units of \(10^{-7}\,\text{N m}\).
On 21 November 1963 a small rocket rose from a fishing village near Thiruvananthapuram. The site, Thumba, was chosen because the magnetic equator passes very close to it. Above the magnetic equator a strong electric current flows in the upper atmosphere, about a hundred kilometres up. Scientists wanted rockets to fly through it and measure it. The Thumba Equatorial Rocket Launching Station began India's space programme at a place of zero dip.
William Gilbert, a physician to Queen Elizabeth I, carved a lodestone into a small sphere. He carried a tiny needle over its surface. Near its poles the needle stood upright; near its middle it lay flat. Sailors had seen compass needles dip in the same way at different latitudes. In his book De Magnete of 1600, Gilbert concluded that the Earth itself is a great magnet.
- Taking \(B_H = B\sin\delta\).The dip is measured from the horizontal, so \(B_H = B\cos\delta\) and \(B_V = B\sin\delta\).
- Calling the magnetic pole in the Arctic a north pole of the Earth's magnet.It attracts the N end of a needle, so it is a south pole of the Earth's magnet.
- Expecting the apparent dip to be smaller than the true dip.Only \(B_H\cos\theta\) lies in a turned plane, so \(\tan\delta' = \tan\delta/\cos\theta\) is larger.
- Thinking a compass points along the total field.A compass turns in the horizontal plane and responds to \(B_H\) only.