Two fields on one table
Lay a bar magnet on a table and put a small compass near it. Two horizontal fields act on the needle: the magnet's field and the Earth's \(B_H\). The needle sets along their vector sum. Close to the magnet, the magnet's field wins. Far away, \(B_H\) wins. In between there can be places where the two are equal and opposite.
A place where the total horizontal field is zero is a neutral point. A compass placed there has nothing to turn it and points in any direction. The magnet's field must be equal to \(B_H\) in size and point south, against \(B_H\). Which places satisfy that depends on how the magnet lies.
A short bar magnet lies along the magnetic meridian with its N pole pointing north. Where are the neutral points?
Choose the chip "N pole north" in the figure above. On the east-west line the field lines of the magnet and of the Earth split apart at P and P′. With a moment of \(3.2\,\text{A m}^2\) in the sheet's field of \(4 \times 10^{-5}\,\text{T}\), they sit about \(20\,\text{cm}\) from the centre. Now choose "N pole south" with a moment of \(1.6\,\text{A m}^2\). The neutral points move onto the axis, again about \(20\,\text{cm}\) away. The drawn magnet has its poles \(5\,\text{cm}\) apart, so the readout is \(19.8\) or \(20.2\,\text{cm}\) rather than exactly \(20\).
A short magnet of moment \(3.2\,\text{A m}^2\) lies along the meridian with its N pole to the north, where \(B_H = 4 \times 10^{-5}\,\text{T}\). Find the neutral points. Then a magnet of \(1.6\,\text{A m}^2\) is laid with its N pole to the south. Find its neutral points.
N pole north: the magnet's field points south on the equatorial line and north on the axis. So the neutral points are on the equatorial line.
\(10^{-7} \times 3.2/d^3 = 4 \times 10^{-5}\), so \(d^3\) \(= 3.2 \times 10^{-7}/4 \times 10^{-5}\) \(= 8 \times 10^{-3}\,\text{m}^3\).
\(d = 0.2\,\text{m}\), east and west of the centre. For the second magnet the axial field points south beyond each pole: \(10^{-7} \times 2 \times 1.6/d^3 = 4 \times 10^{-5}\), so again \(d^3 = 8 \times 10^{-3}\) and \(d = 0.2\,\text{m}\).
Both magnets have neutral points \(0.2\,\text{m}\) from the centre: the first on the east-west line, the second on the north-south axis.
In Delhi, \(B_H = 3.6 \times 10^{-5}\,\text{T}\). A short magnet of moment \(1.8\,\text{A m}^2\) lies along the meridian with its N pole to the north. How far from its centre are the neutral points, in centimetres?
A bar magnet lies along the meridian with its N pole pointing south. Where are the neutral points?
The tangent law
Now turn the magnet so that its field at the needle is perpendicular to \(B_H\), not against it. The needle then feels \(B\) east-west and \(B_H\) north-south. It sets along their resultant, at an angle \(\theta\) from north. From the right-angled triangle of the two fields, \(\tan\theta = B/B_H\). This is the tangent law.
Let a magnetic needle turn freely in a horizontal plane, in two uniform fields \(B\) and \(B_H\) at right angles. It comes to rest at an angle \(\theta\) from \(B_H\) such that \(B = B_H\tan\theta\).
A deflection magnetometer uses the law. A short needle is pivoted at the centre of a circular box. A long light pointer fixed across it moves over a degree scale. Two long arms, graduated in centimetres, carry the magnet. The box is turned until the pointer reads zero, so the needle lies in the meridian. The magnet is then placed on an arm, and the pointer reads \(\theta\).
There are two ways to lay it out. In the tan A position the arms lie east-west and the magnet lies along the arm. The needle is then on the magnet's axis, where the field is along the arm, perpendicular to \(B_H\). In the tan B position the arms lie north-south and the magnet lies across the arm, east-west. The needle is then on the magnet's equatorial line, where the field is again east-west.
At \(20\,\text{cm}\) in tan A the magnet's field is \(4 \times 10^{-5}\,\text{T}\), equal to \(B_H\), and the needle turns \(45°\). In tan B at the same distance the field is half as large, and the needle turns only \(26.6°\). Slide the magnet out and the deflection falls fast, because \(\tan\theta\) goes as \(1/d^3\). To give \(45°\) in tan B the magnet must come in to \(20 \times 2^{-1/3} = 15.9\,\text{cm}\). That is \(0.79\) of the tan A distance, not half of it.
A short magnet, placed \(0.2\,\text{m}\) from the needle, deflects it by \(45°\). Take \(B_H = 4 \times 10^{-5}\,\text{T}\). Find the moment (a) in the tan A position and (b) in the tan B position. (c) Why do the answers differ?
(a) \(M\) \(= B_H\tan\theta\,d^3/(2 \times 10^{-7})\) \(= 4 \times 10^{-5} \times 1 \times 8 \times 10^{-3}/(2 \times 10^{-7})\) \(= 1.6\,\text{A m}^2\).
(b) \(M = B_H\tan\theta\,d^3/10^{-7} = 3.2\,\text{A m}^2\).
(c) The equatorial field is half the axial field at the same distance. So a tan B reading needs twice the moment to give the same deflection.
Values used: \(\mu_0/4\pi = 10^{-7}\,\text{T m/A}\) and the assignment sheet's \(B_H = 4 \times 10^{-5}\,\text{T}\).
For a magnet that is not short, the pole model gives the exact fields of Lesson 1. In tan A, \(\dfrac{\mu_0}{4\pi}\dfrac{2Md}{(d^2 - l^2)^2} = B_H\tan\theta\); in tan B, \(\dfrac{\mu_0}{4\pi}\dfrac{M}{(d^2 + l^2)^{3/2}} = B_H\tan\theta\). A magnet \(10\,\text{cm}\) long at \(20\,\text{cm}\) in tan A gives a field \(14\%\) above the short-magnet value. That is why the practical keeps the magnet well away from the needle.
In Delhi, \(B_H = 3.6 \times 10^{-5}\,\text{T}\). A short magnet of moment \(0.9\,\text{A m}^2\) is placed \(15\,\text{cm}\) from the needle of a magnetometer in the tan A position. Find the deflection, in degrees.
Comparing two magnets
The magnetometer compares two magnets without knowing \(B_H\). In the equal-distance method, both magnets are placed at the same \(d\) in turn. The tan A balance gives \(M \propto \tan\theta\), so \(M_1/M_2 = \tan\theta_1/\tan\theta_2\). In the equal-deflection method, each magnet is moved until the needle shows the same angle. Then \(M/d^3\) is the same for both, so \(M_1/M_2 = d_1^3/d_2^3\).
In the null method, the two magnets go on opposite arms, set so that their fields at the needle point opposite ways. One magnet is moved until the needle returns to zero. Then the two fields cancel, \(M_1/d_1^3 = M_2/d_2^3\), and again \(M_1/M_2 = d_1^3/d_2^3\). In the practical, each reading is taken with the magnet on both arms and with the magnet reversed, and the four are averaged. That removes errors from a scale set slightly off zero and from poles not placed symmetrically.
A magnet A gives a deflection of \(40°\) when \(30\,\text{cm}\) from the needle of a magnetometer in tan A. Magnet B gives the same \(40°\) at \(20\,\text{cm}\). Compare their moments.
Equal deflections mean equal fields at the needle, so \(M_A/d_A^3 = M_B/d_B^3\).
\(M_A/M_B\) \(= (30/20)^3\) \(= 1.5^3\) \(= 3.375\). Magnet A is about three and a third times as strong, which is why it must stand farther away.
Two short magnets are placed in turn at the same distance from the needle of a magnetometer in the tan A position. They give deflections of \(60°\) and \(30°\). Find the ratio of their moments \(M_1/M_2\).
Which statements about the deflection magnetometer are true? Choose all that apply.
- Putting the neutral points on the axis when the N pole points north.On the axis the field is along \(\vec M\); with N to the north it points north and adds to \(B_H\). Look on the equatorial line.
- Using the axial formula for a tan B reading.In tan B the needle is on the equatorial line, so the field is \((\mu_0/4\pi)M/d^3\), without the 2.
- Comparing moments by the ratio of the angles.At equal distances \(M \propto \tan\theta\); at equal deflections \(M \propto d^3\).
- Measuring \(d\) to the near end of the magnet.\(d\) runs from the centre of the magnet to the needle.