Torque on Current Loop, Magnetic Dipole

Case 1: field in the plane of the loop

• Take a rectangular loop $ABCD$ of sides $a$ and $b$, carrying a steady current $I$, placed in a uniform field $\vec{B}$ that lies in the plane of the loop.
• The arms $AD$ and $BC$ are parallel to $\vec{B}$, so they feel no force.
• The arm $AB$ is perpendicular to $\vec{B}$ and feels $F_1 = IbB$ directed into the plane of the loop; the arm $CD$ feels $F_2 = IbB = F_1$ directed out of the plane.
• The two forces are equal, opposite and not collinear: the net force is zero but there is a couple. Taking moments about the central axis, each force acting at $a/2$ from it, $$\tau = F_1\dfrac{a}{2}+F_2\dfrac{a}{2} = IbB\,a = I A B, \qquad A = ab$$
• The torque tends to rotate the loop about the axis joining the midpoints of $AD$ and $BC$.

Case 2: the general orientation

• Now let the normal to the coil make an angle $\theta$ with $\vec{B}$ (the previous case is $\theta = 90^\circ$).
• The forces on $BC$ and $DA$ are equal, opposite and collinear along the coil's axis, so they cancel entirely — no net force and no torque from them.
• The forces on $AB$ and $CD$ are still $F_1 = F_2 = IbB$, equal and opposite but *not* collinear, so they form a couple. The perpendicular distance between them is now $a\sin\theta$: $$\tau = F_1\dfrac{a}{2}\sin\theta + F_2\dfrac{a}{2}\sin\theta = I\,ab\,B\sin\theta = I A B\sin\theta$$
• As $\theta\to 0$ the arms of the couple close up, the forces become collinear, and both net force and torque vanish.

Magnetic moment and the vector form

• Define the magnetic moment of the current loop $$\vec{m} = I\vec{A}, \qquad \vec{m} = N I \vec{A} \text{ for } N \text{ closely wound turns}$$ where the direction of $\vec{A}$ is given by the right-hand thumb rule: curl the fingers of the right hand along the current and the thumb gives $\vec{A}$ (and $\vec{m}$).
• Both cases are then contained in one expression $$\vec{\tau} = \vec{m}\times\vec{B}, \qquad \tau = mB\sin\theta = NIAB\sin\theta$$
• Dimensions of $m$ are $[\text{A}][\text{L}^{2}]$, unit $\text{A m}^{2}$ (also written $\text{J T}^{-1}$).
• This is the exact analogue of the electric dipole in a uniform field, $\vec{\tau} = \vec{p}_e\times\vec{E}$.
• Note the operation: torque is a *cross* product. $-\vec{m}\cdot\vec{B}$, the dot product, is the potential energy, not the torque.

Equilibrium

• $\tau = 0$ when $\vec{m}$ is parallel or antiparallel to $\vec{B}$.
• $\vec{m}\parallel\vec{B}$ ($\theta = 0^\circ$, plane of the loop perpendicular to $\vec{B}$) is *stable*: any small rotation produces a restoring torque. Potential energy $U = -\vec{m}\cdot\vec{B}$ is a minimum here, and the flux of the total field (external plus the loop's own) through the loop is a maximum.
• $\vec{m}$ antiparallel to $\vec{B}$ ($\theta = 180^\circ$) is *unstable*: $U$ is a maximum and any rotation produces a torque that grows with the rotation.
• This is why a small magnet or any magnetic dipole aligns itself with an external field.
• $\theta = 90^\circ$ (the field lying in the plane of the loop) gives the maximum torque $NIAB$.

Points NCERT makes explicitly

• A horizontal current loop on a smooth horizontal plane cannot be made to spin about its own vertical axis by any uniform $\vec{B}$: the area vector $\vec{A}$ is vertical, so $\vec{\tau} = I\vec{A}\times\vec{B}$ always lies in the horizontal plane of the loop, never along the vertical.
• A flexible loop of irregular shape in an external field takes up a circular shape with its plane normal to $\vec{B}$, because for a given perimeter a circle encloses the greatest area and so maximises the flux.
• Consequence for the comparison question: from a wire of length $L$, a square has area $L^{2}/16$ and a circle has area $L^{2}/4\pi$. Since $4\pi\approx 12.57 < 16$, the circle has the larger area, hence the larger magnetic moment and the larger torque. The ratio of moments square to circle is $\pi : 4$.
• Doubling the radius of a loop multiplies its area, and so its magnetic moment $m = IA$, by four.
• The magnetic moment depends on $N$, $A$ and $I$ only — never on the material of the wire.

Circular current loop as a magnetic dipole

• Start from the axial field $B = \dfrac{\mu_0 I R^{2}}{2\left(x^{2}+R^{2}\right)^{3/2}}$ and take $x\gg R$, dropping $R^{2}$ from the denominator: $$B \simeq \dfrac{\mu_0 I R^{2}}{2x^{3}} = \dfrac{\mu_0 I A}{2\pi x^{3}} = \dfrac{\mu_0}{4\pi}\cdot\dfrac{2m}{x^{3}}, \qquad m = IA$$
• Compare the electric dipole's axial field $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2p_e}{x^{3}}$: the two go into each other under the substitutions $\dfrac{1}{\varepsilon_0}\to\mu_0$, $\vec{p}_e\to\vec{m}$, $\vec{E}\to\vec{B}$.
• The same substitutions applied to the electric dipole's equatorial field give, for a point in the plane of the loop at large distance $x$, $$B \simeq \dfrac{\mu_0}{4\pi}\cdot\dfrac{m}{x^{3}} \qquad (x\gg R)$$ — half the axial value.
• These become exact for a point magnetic dipole, and they hold for any planar loop: a planar current loop of $N$ turns is equivalent to a magnetic dipole of moment $\vec{m} = NI\vec{A}$.
• A fundamental difference from the electric case: an electric dipole is built from two monopoles (charges), but in magnetism the dipole (the current loop) is itself the elementary object — magnetic monopoles are not known to exist. This led Ampere to suggest that all magnetism is due to circulating currents; that is partly true, though electrons and protons also carry an intrinsic moment not accounted for by circulating currents.

Diagram in words

• Draw the rectangle $ABCD$ with the current arrows going round it, $\vec{B}$ as horizontal arrows lying in the plane of the loop, $F_1$ into the page on $AB$ and $F_2$ out of the page on $CD$, and $\vec{m}$ perpendicular to the loop.
• Beside it draw the view from the $AD$ end: the loop seen edge-on as a line at angle, the two forces as a couple at its ends, and the perpendicular distance $a\sin\theta$ between their lines of action. Mark $\theta$ between $\vec{m}$ and $\vec{B}$.

Worked numericals

• Square coil, area $100\ \text{cm}^{2} = 10^{-2}\ \text{m}^{2}$, $N = 200$, $I = 5\ \text{A}$, $B = 0.2\ \text{T}$, plane of the coil at $60^\circ$ to the field, so the normal is at $\theta = 90^\circ - 60^\circ = 30^\circ$: $$\tau = NIAB\sin\theta = 200\times5\times10^{-2}\times0.2\times\sin 30^\circ = 1\ \text{N m}$$ Stable equilibrium is at $\theta = 0^\circ$, i.e. with the plane of the coil perpendicular to $\vec{B}$.
• Solenoid of $2000$ turns, $A = 1.6\times10^{-4}\ \text{m}^{2}$, $I = 4.0\ \text{A}$: $m = NIA = 1.28\ \text{A m}^{2}$; in $B = 7.5\times10^{-2}\ \text{T}$ at $30^\circ$ to the axis, $\tau = mB\sin 30^\circ = 4.8\times10^{-2}\ \text{N m}$.
• NCERT: $100$-turn coil of radius $10\ \text{cm}$, $I = 3.2\ \text{A}$ gives $B_{\text{centre}} = \dfrac{\mu_0 NI}{2R} = 2\times10^{-3}\ \text{T}$ and $m = NI\pi r^{2} = 10\ \text{A m}^{2}$; in $B = 2\ \text{T}$ the torque is zero at $\theta=0$ and $mB = 20\ \text{N m}$ at $\theta = 90^\circ$.

Results to carry away

• $\vec{m} = NI\vec{A}$; $\vec{\tau} = \vec{m}\times\vec{B}$; $\tau = NIAB\sin\theta$ with $\theta$ from the *normal*.
• Net force on a current loop in a uniform field is zero.
• $U = -\vec{m}\cdot\vec{B}$: minimum (stable) at $\theta=0^\circ$, maximum (unstable) at $\theta=180^\circ$.
• Far field: axial $\dfrac{\mu_0}{4\pi}\dfrac{2m}{x^{3}}$, equatorial $\dfrac{\mu_0}{4\pi}\dfrac{m}{x^{3}}$.