A magnet that swings
Hang a bar magnet by an untwisted thread so that it can turn in a horizontal plane. It settles along the magnetic meridian. Turn it a little and let go, and it swings to and fro about the meridian. A vibration magnetometer is just this: a magnet on a thread in a glass box that keeps out draughts. A mark on the box helps to count the swings.
The swing is angular SHM, as for the coil in the chapter Magnetic Force on Currents and the Magnetic Dipole. Turned by \(\theta\) from the meridian, the magnet feels a torque \(MB_H\sin\theta\) back towards it. For small angles \(\sin\theta \approx \theta\), so the torque is \(-MB_H\theta\). That is a torsion spring with constant \(\kappa = MB_H\).
\[ \tau = -MB_H\sin\theta \]
Why. the torque on a dipole in a field turns \(\vec M\) back towards \(\vec B_H\).
\[ \tau \approx -MB_H\theta \]
Why. for small angles in radians, \(\sin\theta \approx \theta\).
?
Your turn. Newton's second law for rotation, \(I\alpha = \tau\), with \(I\) the moment of inertia about the thread.
\[ T = 2\pi\sqrt{\dfrac{I}{MB_H}} \]
Read it. this is angular SHM with \(\omega^2 = MB_H/I\), and \(T = 2\pi/\omega\).
For a bar magnet of mass \(m_0\), length \(L\) and breadth \(b\), swinging about a vertical axis through its centre, \(I = m_0(L^2 + b^2)/12\). A stronger magnet, or a stronger field, gives a stiffer spring and a shorter period. A heavier or longer magnet swings more slowly.
(a) A magnet of moment \(1\,\text{A m}^2\) and moment of inertia \(4 \times 10^{-6}\,\text{kg m}^2\) swings in a field \(B_H = 4 \times 10^{-5}\,\text{T}\). Find its period. (b) A magnet of mass \(20\,\text{g}\), \(5\,\text{cm}\) long and \(1\,\text{cm}\) broad, also of moment \(1\,\text{A m}^2\), swings in Delhi. Find its moment of inertia and its period.
(a) \(T\) \(= 2\pi\sqrt{4 \times 10^{-6}/(1 \times 4 \times 10^{-5})}\) \(= 2\pi\sqrt{0.1}\) \(= 1.99\,\text{s}\), about \(2\,\text{s}\).
(b) \(I\) \(= 0.02 \times (0.05^2 + 0.01^2)/12\) \(= 0.02 \times 0.0026/12\) \(= 4.33 \times 10^{-6}\,\text{kg m}^2\).
\(T\) \(= 2\pi\sqrt{4.33 \times 10^{-6}/(1 \times 3.6 \times 10^{-5})}\) \(= 2\pi \times 0.347\) \(= 2.18\,\text{s}\).
Values used: Delhi \(B_H = 3.6 \times 10^{-5}\,\text{T}\); part (a) takes the assignment sheet's \(4 \times 10^{-5}\,\text{T}\).
Two magnets of the same size and mass hang side by side in the same field. The right one has four times the moment of the left one. How does its period compare?
The left magnet of the figure above has \(M = 1\,\text{A m}^2\) and \(I = 4 \times 10^{-6}\,\text{kg m}^2\) in \(4 \times 10^{-5}\,\text{T}\). Its period is \(1.99\,\text{s}\). The right one, with four times the moment, has a period of \(0.99\,\text{s}\). In the time the left magnet makes two swings, the right one makes four.
A magnet of moment \(2\,\text{A m}^2\) and moment of inertia \(8 \times 10^{-6}\,\text{kg m}^2\) swings through small angles in Delhi, where \(B_H = 3.6 \times 10^{-5}\,\text{T}\). Find its period, in seconds.
Periods compare moments and fields
Square the period: \(T^2 = 4\pi^2 I/MB_H\). For two magnets of the same moment of inertia in the same field, \(M \propto 1/T^2\). So \(M_1/M_2 = T_2^2/T_1^2\). Magnets with periods of \(2\,\text{s}\) and \(1\,\text{s}\) have moments in the ratio \(1 : 4\). The faster magnet is the stronger one.
The same magnet can compare two fields. For one magnet, \(B \propto 1/T^2\). A magnet with a period of \(3\,\text{s}\) at one place and \(2\,\text{s}\) at another shows that the second field is \(9/4 = 2.25\) times the first. A magnet placed near the swinging magnet adds its own field, and the period shows that too.
A thin magnet swings with a period \(T\). It is cut into two equal halves (a) across its length, (b) along its length. Find the period of one half in the same field.
(a) Across: each half has moment \(M/2\), half the mass and half the length. For a thin bar \(I \approx m_0L^2/12\), so \(I\) falls to \((1/2)(1/2)^2 = 1/8\) of its value. \(T' = T\sqrt{(1/8)/(1/2)} = T/2\).
(b) Along: each half has moment \(M/2\), half the mass and the full length, so \(I/2\). \(T' = T\sqrt{(1/2)/(1/2)} = T\), unchanged.
A thin bar magnet has a period of \(4\,\text{s}\) in a vibration magnetometer. It is cut across its length into two equal halves. What is the period of one half in the same field?
Two magnets tied together
Two magnets of different sizes have different moments of inertia, so their separate periods do not compare their moments. The sum and difference method gets round this. Tie the two magnets together, one on top of the other, and hang them as one. In the sum position like poles point the same way, and the moment is \(M_1 + M_2\). In the difference position one magnet is turned round, and the moment is \(M_1 - M_2\), with \(M_1\) the larger.
The moment of inertia is \(I_1 + I_2\) in both positions, so it cancels when the two periods are divided. From \(T^2 \propto 1/M\), \((M_1 + M_2)/(M_1 - M_2)\) \(= T_d^2/T_s^2\). With \(T_s = 2\,\text{s}\) and \(T_d = 4\,\text{s}\), as in Fig 4.1, the ratio is \(16/4 = 4\). So \(M_1 + M_2 = 4M_1 - 4M_2\), that is \(3M_1 = 5M_2\) and \(M_1/M_2 = 5/3\).
Two magnets tied together with like poles side by side swing with a period of \(3\,\text{s}\). With one of them reversed the period is \(5\,\text{s}\). Find the ratio \(M_1/M_2\) of the larger moment to the smaller.
Measuring the Earth's field
Neither magnetometer alone gives \(B_H\). The tan A reading gives the ratio \(M/B_H = d^3\tan\theta/(2 \times 10^{-7})\). The period of the same magnet gives the product \(MB_H = 4\pi^2 I/T^2\). Divide the product by the ratio and \(M\) cancels, leaving \(B_H^2\). Multiply them and \(B_H\) cancels, leaving \(M^2\). This is the Class 12 practical way to measure the Earth's field.
In Delhi a short magnet in the tan A position deflects the needle by \(45°\) at \(0.2\,\text{m}\). In a vibration magnetometer the same magnet, of moment of inertia \(5.25 \times 10^{-6}\,\text{kg m}^2\), has a period of \(2.00\,\text{s}\). Find \(B_H\) and \(M\).
Deflection: \(M/B_H\) \(= d^3\tan\theta/(2 \times 10^{-7})\) \(= 8 \times 10^{-3} \times 1/(2 \times 10^{-7})\) \(= 4 \times 10^4\,\text{A m}^2\text{/T}\).
Vibration: \(MB_H\) \(= 4\pi^2 I/T^2\) \(= 39.48 \times 5.25 \times 10^{-6}/4\) \(= 5.18 \times 10^{-5}\,\text{N m}\).
\(B_H\) \(= \sqrt{5.18 \times 10^{-5}/4 \times 10^4}\) \(= \sqrt{1.30 \times 10^{-9}}\) \(= 3.6 \times 10^{-5}\,\text{T}\), and \(M\) \(= \sqrt{5.18 \times 10^{-5} \times 4 \times 10^4}\) \(= 1.44\,\text{A m}^2\).
A student finds \(M/B_H = 5 \times 10^4\,\text{A m}^2\text{/T}\) with a deflection magnetometer and \(MB_H = 8 \times 10^{-5}\,\text{N m}\) with a vibration magnetometer, using the same magnet. Find \(B_H\), in units of \(10^{-5}\,\text{T}\).
Why does the sum and difference method not need the moments of inertia of the two magnets?
The two magnets are tied together in both positions, so the moment of inertia of the pair is the same, \(I_1 + I_2\), in both. The period squared goes as that moment of inertia divided by the net magnetic moment. Dividing the two squared periods cancels the moment of inertia, leaving only \((M_1 + M_2)/(M_1 - M_2)\).
Before 1832, the strength of the Earth's field could only be compared from place to place, in arbitrary units. At Göttingen, Carl Friedrich Gauss hung a magnet and timed its swings, which gave the product of its moment and the field. He then used the same magnet to deflect a second needle, which gave their ratio. Together they gave the field alone, in units of length, mass and time. The method in this lesson is his.
Since 1841 India has recorded the Earth's field near Mumbai, first at Colaba. In the early 1900s Bombay was about to run its trams on electricity, and their currents would disturb the needles. So in 1904 the instruments moved across the harbour to Alibag, away from the city. The Alibag observatory still records the declination, the dip and the horizontal field every day, continuing a record that began in 1841.
- Taking the period to grow with the moment.\(T = 2\pi\sqrt{I/MB_H}\): a larger moment makes a stiffer spring and a shorter period.
- Comparing two different magnets' moments by \(T_2^2/T_1^2\) when their moments of inertia differ.That ratio holds only for equal \(I\). Otherwise tie them together and use the sum and difference method.
- Giving the longer period to the sum position.The sum position has the larger moment, so it swings faster. The longer period belongs to the difference position.