Torque on a Current Loop

When a loop of wire carrying an electric current is placed in a magnetic field, it feels a twisting force, known as torque. This happens because the magnetic field interacts with the moving charges (the electric current) in the wire. This concept is an important part of electromagnetism and is the basic principle behind how devices like electric motors and galvanometers work. The magnetic field pulls on different parts of the loop in different directions, which can cause the loop to rotate. This rotation tends to align the loop so that its surface faces directly into the magnetic field. The amount and direction of the torque depend on several factors: how strong the current is, the size of the loop, the strength of the magnetic field, and the angle between the loop and the field. We can describe this behavior using something called the magnetic dipole moment, which works in a way similar to how electric dipoles behave in electric fields. The torque is strongest when the loop is sideways to the magnetic field and becomes zero when the loop is already aligned with the field.

1.0Definition of Torque on A Current Loop

• Torque on a current loop is the rotational force experienced by a current-carrying loop placed in a magnetic field. This torque arises due to the magnetic forces acting on different segments of the loop, which together create a turning effect that tends to align the loop's plane perpendicular to the magnetic field.

$\vec{\tau }=NI(\vec{A}\times \vec{B})=NIAB\sin \theta$

2.0Diagram of Torque on A Current Loop

3.0Derivation of Torque on A Current Loop

• When a current-carrying coil is placed in a uniform magnetic field, the net magnetic force acting on the coil is always zero.
• However, different segments of the coil may experience magnetic forces in different directions, resulting in a net torque (or couple).
• This torque depends on the orientation of the coil and the axis about which it is allowed to rotate.
• To understand this, consider a rectangular coil placed in a uniform magnetic field $\vec{B}$.
• The coil is free to rotate about a vertical axis PQ, which is normal to the plane of the coil.
• Let the plane of the coil make an angle θ with the direction of the magnetic field
• Under these conditions, the magnetic forces on opposite sides of the coil create a rotational effect (torque) that tends to rotate the coil and align it with the magnetic field.
• The arms AB and CD will experience forces B( NI )b vertically up and down respectively. These two forces together will give zero net force and zero torque (as are collinear with axis of rotation), so will have no effect on the motion of the coil.
• Now the forces on the arms AC and BD will be BINL in the direction out of the page and into the page respectively, resulting in zero net force, but an anticlockwise couple of value,

$\tau =F\times \text{Arm}=BINL\times (b\sin \theta )$

$\tau =BIA\sin \theta \text{ with }A=NlB\dots \dots \dots \dots (1)$

Now treating the current–carrying coil as a dipole of moment $\vec{M}=I\vec{A}Eqn.(1)$ can be written in vector form as

$\vec{\tau }=\vec{M}\times \vec{B}\text{ with }\vec{M}=I\vec{A}=NIA\vec{n}\dots \dots \dots \dots (2)$

Key Result and Its Implications

• Torque will be minimum (= 0) when $\sin \theta =\min =0(i.e.,\theta =0^{\circ }or180^{\circ })$, i.e., the plane of the coil is perpendicular to the magnetic field i.e. normal to the coil is collinear with the field.
• Torque will be maximum (= BINA) when $\sin \theta =\max =1(i.e.,\theta =90^{\circ })$ i.e. the plane of the coil is parallel to the field i.e. normal to the coil is perpendicular to the field.
• By analogy with dielectric or magnetic dipole in a field, in case of current–carrying in a field.

$U=-\vec{M}\cdot \vec{B}withF=-\frac{dU}{dr}$

$W=MB(1-\cos \theta )$

4.0Orientations of The Coil

1.Stable Equilibrium:If $\vec{M}$ is parallel to $\vec{B},\theta =0^{\circ },\tau$ = 0 = Minimum, W = 0 = Minimum, U = -MB = Minimum

 2. No Equilibrium:

$\theta =90^{\circ }(\vec{M}isPerpendicularto\vec{B})$

$\tau =MB=\text{Maximum},W=MB,U=0$

3. Unstable Equilibrium: $\theta =180^{\circ }(\vec{M}isAntiparallelto\vec{B})$

$\tau =0,W=2MB=\text{Maximum},U=MB=\text{Maximum}$

4. Instruments such as electric motor, moving coil galvanometer and tangent galvanometers etc. are based on the fact that a current–carrying coil in a uniform magnetic field experiences a torque (or couple).

5.0Applications of Torque on A Current Loop

1. Electric Motors – Converts electrical energy to mechanical rotation.
2. Galvanometers – Detects small electric currents via coil deflection.
3. Ammeters/Voltmeters – Measures current/voltage using coil torque.
4. Loudspeakers – Produces sound through coil vibration in the magnetic field.
5. Torque Sensors – Measures mechanical torque using magnetic interaction.

Illustration-1.A conducting ring of mass $m=2\text{ kg}$ and radius $R=0.5\text{ m}$,is placed vertically on a smooth horizontal surface, meaning it can rotate without friction. The ring carries a steady current of $I=\frac{1}{\pi }\text{ A}$. At time t = 0 a uniform horizontal magnetic field of magnitude ​​B = 12 T is switched on.If the initial angular acceleration of the ring is given as $\alpha =4x{\text{ rad/s}}^2$, find the value of x

Solution:

${\tau }_{y-\text{axis}}=I_{y-\text{axis}}\alpha$

$(I\pi r^2)B=\frac{1}{2}m{r}^2\alpha$

$\alpha =12{\text{ rad/sec}}^2$

$\alpha =4x{\text{ rad/s}}^2$

$4x=12⟹\therefore x=3$

Illustration-2.A current carrying ring, carrying a constant current $\frac{2}{\pi }\text{ A}radius1m,mass\frac{2}{3}\text{ kg}$ and having 10 windings are free to rotate about its tangential vertical axis. A uniform magnetic field of 1 tesla is applied perpendicular to its plane. How much minimum angular velocity (in rad/sec) should be given to the ring in the direction shown, so that it can rotate 270° in that direction. Write your answer in the nearest single digit in rad/sec.

Solution:

To reach $\theta =270^{\circ }$, it has to cross the potential energy barrier at $\theta =180^{\circ }$ and to cross $\theta =180^{\circ }$ angular velocity at $\theta =180^{\circ }$ should be $0^{+}$

$k_i+U_i=k_f+U_f$

$\frac{1}{2}\left(\frac{3}{2}M{R}^2\right){\omega }^2+(-NIAB\cos 0^{\circ })=0^{+}(-NIAB\cos 180^{\circ })$

$\omega =\sqrt{80}\approx 9\text{ rad/sec}$