KAHIBARO
Table of Contents
Magnetic turning effect on a loop
A current loop placed in a magnetic field can experience a turning effect, called torque. This happens because the magnetic field exerts forces on different parts of the loop, and those forces can form a couple that tends to rotate the loop.
This idea is important because it shows how electricity can produce mechanical rotation. Electric motors are based on this effect.
A rectangular loop in a uniform magnetic field
Consider a rectangular loop carrying current $I$ in a uniform magnetic field $\vec{B}$. The loop has sides of lengths $a$ and $b$, so its area is
$$
A = ab
$$
If the magnetic field is parallel to some sides of the loop, those sides feel no magnetic force because for a straight wire the magnetic force is
$$
\vec{F} = I \vec{L} \times \vec{B}
$$
The two opposite sides that are not parallel to the field feel equal forces in opposite directions. These forces do not usually cancel in their rotational effect. Instead, they can create a torque that tends to rotate the loop.
Why the loop rotates
The magnetic forces on opposite sides of the loop are equal in magnitude and opposite in direction, so the net force on the loop can be zero. But zero net force does not mean zero torque.
A pair of equal and opposite forces acting at different points forms a couple. A couple does not translate the object, but it can rotate it. In a current loop, the field pushes one side upward and the opposite side downward, or in similar opposite directions depending on orientation. This produces rotation.
The loop turns so that its plane becomes perpendicular to the magnetic field, or equivalently, so that the loop's area vector aligns with the magnetic field.
Torque magnitude
For a single loop, the torque magnitude is
$$
\tau = IAB \sin \theta
$$
where $I$ is the current, $A$ is the area of the loop, $B$ is the magnetic field magnitude, and $\theta$ is the angle between the loop's area vector and the magnetic field.
If the loop has $N$ turns, the torque becomes
$$
\tau = NIAB \sin \theta
$$
Important formula:
$$
\tau = NIAB \sin \theta
$$
Here, $\theta$ is the angle between $\vec{B}$ and the area vector, not the angle between $\vec{B}$ and the plane of the loop.
Magnetic dipole moment
A current loop behaves like a magnetic dipole. Its magnetic dipole moment is defined as
$$
\vec{\mu} = N I A \hat{n}
$$
where $\hat{n}$ is a unit vector perpendicular to the plane of the loop. Its direction is found using the right hand rule. Curl the fingers of your right hand in the direction of the current, and your thumb points in the direction of $\vec{\mu}$.
Using this quantity, the torque can be written in vector form as
$$
\vec{\tau} = \vec{\mu} \times \vec{B}
$$
and its magnitude is
$$
\tau = \mu B \sin \theta
$$
Key vector relation:
$$
\vec{\tau} = \vec{\mu} \times \vec{B}
$$
The torque tends to rotate the loop so that $\vec{\mu}$ lines up with $\vec{B}$.
The area vector
The area vector is not a vector because area itself points somewhere in space in an obvious physical sense, but it is a very useful mathematical tool. Its magnitude equals the area of the loop, and its direction is perpendicular to the loop.
For a flat loop,
$$
\vec{A} = A \hat{n}
$$
so the magnetic dipole moment may also be written as
$$
\vec{\mu} = N I \vec{A}
$$
This makes the torque formula look compact and clear.
Special cases
The torque depends on the angle $\theta$.
When $\theta = 0^\circ$, the area vector is parallel to the magnetic field. Then
$$
\tau = NIAB \sin 0^\circ = 0
$$
There is no turning effect in this orientation.
When $\theta = 90^\circ$, the area vector is perpendicular to the field. Then
$$
\tau = NIAB \sin 90^\circ = NIAB
$$
This is the maximum torque.
The behavior is summarized below.
| Orientation | Angle $\theta$ between $\vec{\mu}$ and $\vec{B}$ | Torque |
|---|---|---|
| Dipole aligned with field | $0^\circ$ | $0$ |
| Dipole opposite to field | $180^\circ$ | $0$ |
| Dipole perpendicular to field | $90^\circ$ | Maximum |
Rotational tendency and stability
The torque acts to reduce the angle between $\vec{\mu}$ and $\vec{B}$. If the dipole is already aligned with the field, small disturbances do not make it rotate away easily, so this is a stable equilibrium. If the dipole points exactly opposite to the field, the torque is also zero, but that position is unstable. A small disturbance causes rotation toward alignment.
Torque and potential energy
A current loop in a magnetic field also has magnetic potential energy
$$
U = - \vec{\mu} \cdot \vec{B}
$$
or
$$
U = -\mu B \cos \theta
$$
This energy is lowest when $\vec{\mu}$ is parallel to $\vec{B}$. That is why the loop tends to rotate into that orientation.
Important energy relation:
$$
U = - \vec{\mu} \cdot \vec{B}
$$
Minimum energy occurs when the magnetic dipole moment points in the same direction as the magnetic field.
A simple example
Suppose a coil has $N = 20$ turns, each of area $A = 5.0 \times 10^{-3}\,\text{m}^2$, carrying current $I = 2.0\,\text{A}$ in a magnetic field $B = 0.40\,\text{T}$. If the angle between $\vec{\mu}$ and $\vec{B}$ is $30^\circ$, the torque is
$$
\tau = NIAB \sin \theta
$$
$$
\tau = (20)(2.0)(5.0 \times 10^{-3})(0.40)\sin 30^\circ
$$
Since $\sin 30^\circ = 0.5$,
$$
\tau = 20 \times 2.0 \times 5.0 \times 10^{-3} \times 0.40 \times 0.5
$$
$$
\tau = 0.040\,\text{N m}
$$
So the loop experiences a torque of
$$
\tau = 4.0 \times 10^{-2}\,\text{N m}
$$
Use in electric motors
In a motor, a current loop placed in a magnetic field experiences torque and begins to rotate. To keep the rotation going in the same direction, the current is reversed every half turn in a simple DC motor. This keeps the torque from changing sign in a way that would stop the motion.
The physical principle is the torque on current loops.
What to remember
A current loop in a magnetic field can rotate even when the net force is zero. The turning effect comes from a pair of magnetic forces acting on opposite sides of the loop. The loop acts like a magnetic dipole with moment
$$
\vec{\mu} = NIA\hat{n}
$$
and the torque is
$$
\vec{\tau} = \vec{\mu} \times \vec{B}
$$
with magnitude
$$
\tau = NIAB\sin\theta
$$
Essential result for current loops in magnetic fields:
$$
\vec{\tau} = \vec{\mu} \times \vec{B}, \qquad \vec{\mu} = NI\vec{A}
$$
The loop rotates to align its magnetic dipole moment with the external magnetic field.