Torque Acting on a Magnetic Dipole in a Uniform Magnetic Field

Torque on a magnetic dipole

The set-up and the diagram

• Draw a short bar magnet of magnetic dipole moment $\vec{m}$ (arrow pointing from its south pole to its north pole) placed in a uniform magnetic field $\vec{B}$ shown as equally spaced parallel horizontal lines. Let $\theta$ be the angle between $\vec{m}$ and $\vec{B}$.
• The field exerts a force on the north pole along $\vec{B}$ and an equal, opposite force on the south pole. The two forces are equal in magnitude, opposite in direction and act along different lines of action — that is a couple.
• A couple gives zero net force but a non-zero moment, so the magnet undergoes pure rotational motion, not translation.

The result

• In Chapter 10 the torque on a rectangular current-carrying coil in a uniform field was obtained as

$$\vec{\tau} = \vec{m}\times\vec{B}\qquad \tau = mB\sin\theta \qquad \ldots (1)$$

• where $\vec{m}$ is the magnetic dipole moment and $\vec{B}$ the external applied uniform field. A short bar magnet in a uniform field behaves in exactly the same way and obeys the same equation.
• $\tau$ is maximum, equal to $mB$, when $\theta = 90^\circ$, and zero when $\theta = 0^\circ$ or $180^\circ$.
• This is the torque that turns a compass needle back into the north–south line when you force it into the east–west direction and release it.

Magnetic potential energy of the dipole

• Wherever an angular displacement takes place, work is done, and that work is stored as potential energy in the new position. For a magnetic dipole this stored energy is the magnetic potential energy $U_m$, the exact analogue of the electrostatic potential energy of an electric dipole in an electric field.

$$U_m = \int \tau\, d\theta \qquad \ldots (2)$$

$$U_m = \int mB\sin\theta\, d\theta$$

$$\therefore\ U_m = -mB\cos\theta = -\vec{m}\cdot\vec{B} \qquad \ldots (3)$$

The three cases

• Case 1: $\theta = 0^\circ$, $\cos 0^\circ = 1$, so $U_m = -mB$. Here $\vec{m}$ and $\vec{B}$ are parallel. The magnet has minimum potential energy and is in its most stable state.
• Case 2: $\theta = 180^\circ$, $\cos 180^\circ = -1$, so $U_m = +mB$. Here $\vec{m}$ and $\vec{B}$ are antiparallel. The magnet has maximum potential energy and is in its most unstable state.
• Case 3: $\theta = 90^\circ$, $\cos 90^\circ = 0$, so $U_m = 0$. The magnet is perpendicular to the field.
• Diagram: the graph of $U_m$ against $\theta$ is a cosine curve starting at $-mB$ at $\theta = 0^\circ$, crossing zero at $90^\circ$, reaching $+mB$ at $180^\circ$.

Work done in rotation

$$W = U_2 - U_1 = mB(\cos\theta_1 - \cos\theta_2)$$

• Rotating the magnet from its most stable to its most unstable position needs $W = mB(\cos 0^\circ - \cos 180^\circ) = 2mB$.
• Rotating through $90^\circ$ from the magnetic meridian needs $mB$; through $60^\circ$ it needs $\dfrac{mB}{2}$. The first is therefore 2 times the second.

Angular oscillations of a suspended magnet

The set-up and the diagram

• Draw a short bar magnet hanging from a rigid support by an inextensible string, free to turn about the vertical axis of suspension, with the earth's horizontal field (or a uniform field $\vec{B}$) marked along the north–south line.
• Displace the magnet through a small angle and release it. The deflecting torque due to the field is opposed by the restoring torque, the magnet turns through an angle, comes back, and oscillates about the mean position instead of rotating continuously.

Derivation of the period

• The torque producing the angular acceleration is

$$\tau = I\dfrac{d^2\theta}{dt^2} \qquad \ldots (4)$$

• where $I$ is the moment of inertia of the magnet about the axis of suspension and $\dfrac{d^2\theta}{dt^2}$ the angular acceleration.
• The restoring torque acts opposite to the angular displacement:

$$\tau = -mB\sin\theta \qquad \ldots (5)$$

• Equating (4) and (5),

$$I\dfrac{d^2\theta}{dt^2} = -mB\sin\theta$$

• For a small angular displacement, $\sin\theta \approx \theta$, so

$$I\dfrac{d^2\theta}{dt^2} = -mB\theta \qquad\Rightarrow\qquad \dfrac{d^2\theta}{dt^2} = -\left(\dfrac{mB}{I}\right)\theta \qquad \ldots (6)$$

• The angular acceleration is proportional to the angular displacement and directed oppositely, with $m$, $B$ and $I$ constant. This is angular S.H.M., the exact analogue of linear S.H.M. governed by $\dfrac{d^2x}{dt^2} + \omega^2 x = 0$.

$$\therefore\ \omega^2 = \dfrac{mB}{I}\qquad \omega = \sqrt{\dfrac{mB}{I}} \qquad \ldots (7)$$

$$\therefore\ T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{I}{mB}} \qquad \ldots (8)$$

Vibration magnetometer

• Principle: whenever a freely suspended magnet in a uniform magnetic field is disturbed from its equilibrium position, it starts vibrating about the mean position with the period given by Eq. (8).
• It is used to compare magnetic moments of two magnets and to compare magnetic fields, and to determine the horizontal component of the earth's magnetic field.

Location of magnetic poles of a current-carrying loop

• Draw a circular loop on a glass plate and pass a current round it. Looking at the plate from the front with the current appearing clockwise, the near face of the loop behaves as a south pole (right-hand rule).
• Turn the glass over and look at the same loop from the other side. The current now appears anticlockwise, and that face behaves as a north pole. The direction of the current has not changed at all — only the side from which it is being viewed.

Standard results

• $\vec{\tau} = \vec{m}\times\vec{B}$, $\tau = mB\sin\theta$; maximum $mB$ at $90^\circ$, zero at $0^\circ$ and $180^\circ$; net force zero.
• $U_m = -mB\cos\theta$; $U_{min} = -mB$ at $0^\circ$ (most stable), $U_m = 0$ at $90^\circ$, $U_{max} = +mB$ at $180^\circ$ (most unstable).
• $W = mB(\cos\theta_1 - \cos\theta_2)$; stable to unstable costs $2mB$.
• $T = 2\pi\sqrt{\dfrac{I}{mB}}$, hence $m = \dfrac{4\pi^2 I}{T^2 B}$ and $B = \dfrac{4\pi^2 I}{T^2 m}$.
• Breaking the magnet into two equal halves gives $I' = \dfrac{I}{8}$ and $m' = \dfrac{m}{2}$, so $T' = \dfrac{T}{2}$.