Torque on a Current Carrying Loop and Magnetic Dipole:

Torque on a Current Carrying Loop and Magnetic Dipole: The most common use for current-carrying conductors is electric motors and generators. We see different types of electric motors and generators in fans, washing machines, electric mixers, generating stations, industries, etc. motors have found use at different places. Have you ever wondered how exactly do these motors work? What is the basic principle behind it? What exactly makes them rotate at such a high speed?

In this article, we are going to look at the basic concept of its working. We will see how a current-carrying loop rotates in an external applied magnetic field.

Magnetic Dipole

It is analogous to electric dipole from electrostatics. Here, we have two magnetic poles, viz., the north pole and the south pole, separated by a distance. But the change here is that we can observe individual positive and negative electric charges in nature, but in the case of magnetism, magnetic monopoles, i.e., individual north pole or south pole, have never been observed in nature. Hence, we conclude that magnetic monopoles do not exist. They are always present in the form of a magnetic dipole.

The picture here is of two basic magnets that are a bar magnet and a horseshoe-shaped magnet. Here, the north pole and south pole exist together, and they can also be termed as magnetic dipoles.

There are other shaped magnets, too, like disc-shaped, ring-shaped, oval-shaped, etc. These two are the most common magnet which we see. 

Apart from this, a current-carrying conductor also creates a magnetic field around it. The magnetic field on both sides of the conductor is in opposite directions. Hence, here too, we get a magnetic dipole. 

Using the right-hand thumb rule, we can determine the direction of the magnetic field,

Using the right-hand thumb rule, we can determine the direction of the magnetic field, which is in a loop, as shown in the diagram. If we consider the conductor along the plane of paper or your screen, then the magnetic field lines coming out of the plane are generally considered to be the north pole, and those going inside the plane are considered to be the south pole. So from the diagram, we get the north pole on the left side of the conductor and the south pole on the right side of the conductor, and this forms a magnetic dipole.

Also, we can also see a magnetic dipole on a current-carrying loop. The magnetic field is generated inside a current-carrying loop based on the direction of current flowing in the loop. 

From the diagram, determining the direction of the magnetic field using the right-hand thumb rule, magnetic field lines are in the upward direction, which means they are coming out of the plane of the loop. As discussed earlier, if magnetic field lines come out of the plane, they are considered to be the north pole and the field going inside the plane is considered to be the south pole. Thus, on the upper side of the current-carrying loop, the north pole is induced, and on the bottom side south pole is induced, and this again forms a magnetic dipole.

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Magnetic Dipole Moment

In electrostatics, the electric dipole moment is defined as the product of the magnitude of charge and distance between both the charges. Similarly, for a bar magnet, we can define magnetic dipole moment as the product of pole strength \((p)\) of any one of the poles and the magnetic length (effective distance between both the poles) of the bar magnet. 

\(\overrightarrow \mu  \,\, = p\overrightarrow L \)

Pole strength determines the ability of a magnet to create a magnetic field. The direction of magnetic dipole moment inside the magnet is from the south pole to the north pole. 

Apart from this, magnetic dipole moment is also defined for a current-carrying loop. If the loop carries a current I and has a cross-sectional area A, then its magnetic dipole moment is defined to be the product of current and area vector of the cross-sectional area of the loop. 

\(\overrightarrow \mu  \, = I\overrightarrow A \)

So from this equation, it is clear that the direction of the magnetic moment for a

current-carrying loop is along the area vector, i.e. perpendicular to the cross-sectional area of the loop.

From this equation, we can say that the SI unit of magnetic moment is

\(A{m^2}(Ampere – metr{e^2})\)

Magnetic Moment Due to a Revolving Electron in Its Orbit

Let us take the case of an electron revolving around a proton in its fixed orbit, as shown in the diagram. The electron here revolves around a proton in the counterclockwise direction in a fixed orbit with a constant speed v. Hence, the direction of current due to this flow of electrons will be opposite to that of the electron, i.e., in the clockwise direction. Due to this, the magnetic field and hence its magnetic moment will be in the downward direction.

If the time period of its rotation is T and the radius of its circuit is r, the current due to this motion of electron will be

\(I = \frac{e}{T}\)

Since the speed of the electron is constant, we can write for its time period as,

\(T = \frac{{2\pi r}}{v}\)

Hence, the equation of current will be,

\(I = \frac{{ev}}{{2\pi r}}\)

The magnetic moment of a current-carrying loop is given by the equation

\(\mu  = IA\)

\(\mu  = \frac{{ev}}{{2\pi r}}(\pi {r^2})\)

\(\therefore \mu  = \frac{{evr}}{2}\)

Multiplying and dividing this equation by mass of electron \(({m_e})\)

\(\mu  = \frac{{ev}}{{2{m_e}}}({m_e}vr)\)

But \({m_e}vr\) is the angular momentum of the electron in that particular orbit

\(\therefore \mu  = (\frac{e}{{2{m_e}}})L\)

\(\therefore \vec \mu  =  – (\frac{e}{{2{m_e}}})\vec L\)

From the above equation, we can conclude that the magnetic moment of an electron in its orbit depends on the angular momentum of its orbit.

The ratio\(\frac{e}{{2{m_e}}}\) is known as the gyromagnetic ratio, and its value is approximate

\(8.8 \times {10^{10}}{\raise0.7ex\hbox{$C$} \!\mathord{\left/

 {\vphantom {C {{\rm{kg}}}}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{${{\rm{kg}}}$}}\)

According to Bohr’s hypothesis, the angular momentum of an electron in any orbit is the integral multiple of \(\frac{h}{{2\pi }}\) where h is known as Planck’s constant and its value is \(6.626 \times {10^{ – 34}}Js\)

\(\mu  = (\frac{e}{{2{m_e}}})(\frac{{nh}}{{2\pi }})\)

\(\mu  = n(\frac{{eh}}{{4\pi {m_e}}})\)

Here, the ratio\(\frac{{eh}}{{4\pi {m_e}}}\) is known as Bohr Magneton\[(BM)\] and its value is\(9.274 \times {10^{ – 24}}A\,{m^2}\)So, for\(n = 1(first\,orbit)\), we get\(\mu  = 1BM\) This even acts as a unit for the magnetic moment for an electron revolving in its orbit. For example, if we consider the\({3^{rd}}\) orbit, then its magnetic moment will be \(3BM\). Hence, the higher the orbit, the more is its magnetic moment.

Torque Acting on a Current Carrying Loop Placed in a Uniform Magnetic Field

When a conductor carrying a constant current is placed in a uniform magnetic field, a magnetic force acts on it, which is given by the equation 

\(\vec F = I\vec L \times \vec B\)

Here, L is the length of the conductor, and B is the external magnetic field in which this conductor is placed. From this equation, we can say that the force acting on the conductor is always perpendicular to the direction of the magnetic field and the direction of current inside the conductor.

Consider a current-carrying rectangular loop of length l and width w as shown in the diagram. Let us name this loop as PQRS. The direction of current in this loop will be in the direction of P to Q to R to S. 

Force acting on the side PQ of the loop will be given by

\({\vec F_{PQ}} = I\vec L \times \vec B\)

Hence, the direction of the force on this conductor will be in the direction as shown in the diagram. The same will be the magnitude of the force on the conductor RS, but its direction will be opposite to that of PQ. 

The torque acting on both the conductors are depicted clearly in the given diagram. The torque due to both the conductors will be in the same direction (upward) and will be given by the equation

\(\tau  = Il\frac{w}{2}B\,\sin \,\sin \,\theta \)

This equation is for the torque acting on a single conductor.

Hence, the total torque due to both the conductors will be

\({\tau _{net}} = IlwB\,\sin \,\sin \,\theta \)

Talking about conductor QR and PS, the force acting on both will be in the downward and upward direction, respectively. Hence, the torque acting on both the conductors with respect to the centre of the loop will be zero, as shown in the diagram.

Therefore, the net torque on the current-carrying loop will be equal to the net torque on conductors PQ and RS only.

\({\tau _{net}} = IlwB\,\sin \,\sin \,\theta \)

But lw=A (Area of the loop)

\(\therefore {\tau _{net}} = IAB\,\sin \,\sin \,\theta \)

Writing in vector form

\(\therefore {\tau _{net}} = I\vec A\, \times \vec B\)

But the product of current and area is known as a magnetic moment. Therefore, the torque acting on the loop will be given by:

\(\therefore {\tau _{net}} = \vec \mu \, \times \vec B\)

So, the torque acting on a current-carrying loop will depend on its magnetic moment and the external magnetic field in which the loop is placed.

Solved Examples

Q.1. A current\(0.5\,mA\)is set in a loop of cross-sectional area\(10c{m^2}\)What is the magnetic moment for this loop?
Ans: Current in the loop \((I) = 0.5mA\)
The cross-sectional area of the loop \((A) = 10\,c{m^2}\)
The magnetic moment of a current-carrying loop is given by the equation
\(\mu  = IA\)
Substituting the values:
\(\mu  = 0.5 \times {10^{ – 3}} \times 10 \times {10^{ – 4}}\)
\(\therefore \mu  = 0.5 \times {10^{ – 7}}A\,{m^2}\)

Q.2. Find the torque acting on a loop of area \({\rm{40}}\,{\rm{c}}{{\rm{m}}^2}\) carrying a current of \({\rm{0}}{\rm{.1}}\,{\rm{A}}\) placed along an external magnetic field of \({\rm{500}}\,\mu {\rm{T}}\).
Ans: Current in the loop\((I) = 0.1A\)
Area of the loop \((A) = \,40\,{\rm{c}}{{\rm{m}}^2}\)
External magnetic field \((B)\, = \,500\,\mu T\)
Since the loop is along the magnetic field, the angle between its area vector and the direction of magnetic field lines will be \({90^ \circ }\). Hence, the magnitude of torque acting on the loop will be given by
\(\tau \, = IAB\)
Substituting the values:
\(\tau \, = \,0.1 \times 40 \times {10^{ – 4}} \times 500 \times {10^{ – 6}}\)
\(\therefore \tau  = 2 \times {10^{ – 7}}\,{\rm{N}}\,{\rm{m}}\)

Summary

From this article, we got to know a little bit about magnets and magnetic fields, magnetic moment for permanent magnet and current-carrying loop, and finally how does an external magnetic field effect loop carrying current. Mainly the study of magnetic dipole moment is important for studying the effect of magnetic field on a loop or any other magnet because it helps us with torque on the loop and also to determine magnetic energy.

The magnetic moment of a bar magnet: \(\overrightarrow \mu  \, = \,p\overrightarrow L \)

The magnetic moment of a current-carrying loop: \(\overrightarrow \mu  \, = \,I\overrightarrow A \)

The torque acting on a current-carrying loop placed in an external magnetic field: \({\overrightarrow \tau  _{net}}\, = \,\overrightarrow \mu  \,\, \times \,\overrightarrow B \)

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Frequently Asked Questions (FAQs)

Q.1. How do you find the direction of torque on a current loop?
Ans: The direction of torque on a current loop will be the same as the direction of cross-product of area vector and external magnetic field.

Q.2. What is the torque on the current loop placed in a magnetic field?
Ans: The magnitude of the torque on a current loop placed in the external magnetic field will equal its magnetic moment times magnetic field.

Q.3. Under what condition, torque on a current loop is zero?
Ans: The torque acting on a current loop is zero when the magnetic field lines enter perpendicularly in the loop because the angle between the area vector and the magnetic field is zero degrees.

Q.4. What is the SI unit of the magnetic moment?
Ans: The SI unit of magnetic moment is \({\rm{A}}{{\rm{m}}^2}\)

Q.5. Define magnetic dipole moment.
Ans: For a permanent magnet, the magnetic dipole moment is defined ad the product of pole strength and the relevant distance between both the magnetic poles. Whereas for a current-carrying loop, we can define magnetic dipole moment as the product of current in the loop and its area vector.

We hope this detailed article on torque on a current-carrying loop and magnetic dipole helped you in your studies. If you have any doubts, queries or suggestions regarding this article, feel to ask us in the comment section and we will be more than happy to assist you. Happy learning!