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In this Physics tutorial, you will learn:

- What is magnetic moment? How it is produced?
- What is magnetic dipole moment? What type of quantity is it (vector or scalar)?
- How to calculate the magnetic dipole moment?
- How to calculate the magnetic torque in a coil in terms of magnetic dipole moment?
- What are the similarities between magnetic and electric torque produced in a current carrying loop?
- How to calculate the energy of a magnetic dipole?
- How to calculate the work done to rotate a magnetic dipole?

μ = N ∙ I ∙ A

where N is the number of turns in the coil, I is the current in the coil and A is the area of the coil. Obviously, the unit of magnetic dipole moment is [A ∙ m2]. Example: A current carrying coil having a resistance of 30 Ω is connected to a circuit supplied by a 12 V battery as shown in the figure. The diameter of coil is 8 cm. Calculate the magnitude of the magnetic dipole moment produced by the coil. Solution Clues: ε = 12 V R = 30 Ω d = 8 cm = 0.08 m N = 5 (number of turns in the coil as seen in the figure) μ = ? First, we calculate the current flowing in the coil using the Ohm's Law. We have: I = ε/R = 12V/30Ω = 0.4 A

Second, we calculate the area of coil. We have: A = π ∙ (d/2)^{2} = 3.14 ∙ (0.08m/2)^{2} = 3.14 ∙ 〖0.0016m〗^{2} = 0.005 m^{2}

Now, let's calculate the magnitude of magnetic dipole moment. We have μ = N ∙ I ∙ A = 5 ∙ 0.4A ∙ 0.005 m^{2} = 0.01 A ∙ m^{2}

Torque in Terms of Magnetic Dipole Moment In tutorial 16.3 we have seen that the maximum torque τmax produced when a current carrying loop is inserted in a magnetic field produced on the lateral sides of the loop (when it is rectangular-shaped) is τ = I ∙ B ∙ A ∙ sinθ

or τ_max = I ∙ B ∙ A

(this occurs when the angle θ formed by the magnetic field and current is 900, i.e. when sin θ = 1) The approach used in 16.3 can be applied for loops having other shapes as well. We are interested to see what happens with the torque when the loop is circular in order to express it in terms of magnetic dipole moment. Thus, combining the equations μ = N ∙ I ∙ A

and τ = I ∙ B ∙ A ∙ sinθ

we obtain for the magnetic torque in terms of the magnetic dipole moment τ = μ ∙ B ∙ sinθ

The vector form of the above equation is τ*⃗* = μ*⃗* × B*⃗*

which is very similar to the torque produced by an electric field Eτ*⃗* = p*⃗* × E*⃗*

In either case, the torque is the vector product of the specific dipole to the corresponding field. Example: A 0.5A circular current carrying loop has a radius of 2 cm. The loop is inserted inside a 0.2 T magnetic field produced by two bar magnets as shown in the figure. Calculate: a) The magnetic dipole moment produced at the loop b) The torque on the loop produced by the magnetic field Solution a) The magnetic dipole moment on this loop is μ = N ∙ I ∙ A

where N = 1 (there is a single loop in the coil) and A = π ∙ r2 is the area of loop (r = 2 cm = 0.02 m). Hence, we have μ = N ∙ I ∙ π ∙ r^{2} = 1 ∙ (0.5A) ∙ 3.14 ∙ (0.02m)^{2} = 0.000628 A ∙ m^{2} = 6.28 × 10^{-4} A ∙ m^{2}

b) The direction of magnetic dipole moment is found by applying the right hand rule provided in theory (when grasping the magnetic dipole moment with the right hand, the four fingers show the direction of current and thumb shows the direction of magnetic dipole moment). Thus, we find that the magnetic dipole moment vector is directed downwards. On the other hand, the magnetic field lines are directed upwards (from N-pole to S-pole, or from red to blue). Therefore, the angle θ formed by these two vectors is 1800. Hence, we obtain for the torque on the loop: τ = μ ∙ B ∙ sinθ = 6.28 × 10^{-4} A ∙ m^{2} ∙ 0.2T ∙ sin〖180^0 〗 = 6.28 × 10^{-4} A ∙ m^{2} ∙ 0.2T ∙ 0 = 0

This result means no turning effect is produced on the loop. Energy of a Magnetic Dipole As we know from Section 5, objects and systems tend to occupy a state in which they have the lowest energy possible. For example, when you roll a ball inside a hole, it swings several times until it stops at bottom of the hole in which the ball is closer to the ground and therefore it has the lowest gravitational energy. Likewise, when you make a pendulum swing, it will stop at the vertical position, in which the bob is closer to the ground. In this case, the bob has the lowest gravitational potential energy possible. Another example: people lie on the bed when sleeping so that the centre of gravity of the body be as close as possible to the ground. In this case, they have the lowest energy possible and therefore, they can take a rest. This rule is also true when a magnetic dipole is inserted inside an external magnetic field. When the vectors of magnetic field and magnetic dipole moment are collinear, the system has the lowest energy as the magnetic moment vector attempts to align with the magnetic field. Following this reasoning, we can say that when magnetic moment vector is antiparallel to the magnetic field lines (as in the previous solved example), the system has the highest energy possible. Therefore, we can say that a magnetic dipole in an external field has an energy that depends on the orientation of dipole moment in respect to the magnetic field direction. For electric dipoles, we have for the energy U in terms of the angle θ: U(θ) = -p*⃗* ∙ E*⃗*

where pU(θ) = -μ*⃗* ∙ B*⃗*

Note that here we have a dot product of two vectors because energy is a scalar (the dot product of two vectors gives a scalar). The sign minus is because the maximum value of energy is obtained for θ = 1800 for which cos θ = -1. In this way, we obtain a positive value for the maximum energy. Example: Calculate the energy of the magnetic dipole shown in the figure below if the radius of loop is r = 3cm, the current flowing through the loop is I = 5A and the magnetic field lines punch the area of loop at α = 600. The magnitude of magnetic field is B = 40mT. Solution Clues: r = 3 cm = 3 × 10-2 m I = 5A α = 600 B = 40 mT = 4 × 10-2 T N = 1 U = ? The vector of magnetic dipole moment is directed upwards. This is found using the right hand rule mentioned earlier. The angle formed by the magnetic dipole moment vector to the magnetic field lines is θ = 900 + α = 900 + 600 = 1500. The magnitude of magnetic dipole moment μ is μ = N ∙ I ∙ A = N ∙ I ∙ π ∙ r^{2} = 1 ∙ (5A) ∙ 3.14 ∙ (3 × 10^{-2} m)^{2} = 1.413 × 10^{-2} A ∙ m^{2}

The energy of the magnetic dipole therefore is U(θ) = -μ*⃗* ∙ B*⃗* = -μ ∙ B ∙ cosθ = -(1.413 × 10^{-2} A ∙ m^{2} ) ∙ (4 × 10^{-2} T) ∙ (cos〖150^0 〗 ) = -(1.413 × 10^{-2} A ∙ m^{2} ) ∙ (4 × 10^{-2} T) ∙ (-0.866) = 4.89 × 10^{-4} J

Work Done on a Magnetic Dipole If an external force rotates the coil (magnetic dipole) from an angle θ1 to another angle θ2, a resulting torque is produced. This action results in a work W done on the dipole by the applied torque. If the dipole is stationary before and after the change in the angle mentioned above, then the work done on the magnetic dipole is W = U_final-U_initial

There are many more examples of magnetic dipoles besides the circular or rectangular current carrying loops discussed so far. For example, a charged rotating sphere can be a magnetic dipole, subatomic particles such as protons and electrons can be considered as magnetic dipoles as well because they have magnetic dipole moments and so on. All these quantities can be viewed as tiny current carrying loops. Example: The figure below shows a lateral view of a circular coil containing 100 turns. The coil's diameter is 4 cm and a 200mA current is flowing through it. The coil is initially at rest inside a 0.500 T magnetic field at the position 1 shown in the figure. Then it rotates by means of an external force at the position 2. Calculate: a) The direction of electric current flowing through the coil b) The work done by the external force to move the coil from position 1 to position 2 Solution a) Applying the right hand rule where the thumb is in the direction of magnetic dipole moment μ and the four fingers show the direction of current, it is obvious that the current in = b) We have the following clues: θ1 = 00 θ2 = 900 d = 4 cm = 0.04 m I = 200 mA = 0.200 A B = 0.500 T N = 100 W = ? From the equation W = ∆U = U_f-U_i = U(θ_{2} )-U(θ_{1} ) = [-μ ∙ B ∙ cos〖θ_{2} 〗 ]-[-μ ∙ B ∙ cos〖θ_{1} 〗 ] = [-N ∙ I ∙ A ∙ B ∙ cos〖θ_{2} 〗 ]-[-N ∙ I ∙ A ∙ B ∙ cos〖θ_{1} 〗 ] = [-N ∙ I ∙ A ∙ B ∙ cos〖90^0 〗 ]-[-N ∙ I ∙ A ∙ B ∙ cos〖0^0 〗 ] = [-N ∙ I ∙ A ∙ B ∙ 0]-[-N ∙ I ∙ A ∙ B ∙ 1] = N ∙ I ∙ A ∙ B = N ∙ I ∙ π ∙ (d/2)^{2} ∙ B = 100 ∙ (0.200A) ∙ 3.14 ∙ (0.04m/2)^{2} ∙ (0.500T) = 0.01256 J = 12,56 mJ

Summary Magnetic moment is a vector quantity that describes magnetic processes. More specifically, it indicates the strength of a magnetic dipole, which represents the relationship between the external magnetic field and the strength of that field itself. Generally, there can be two origins for the creation of the magnetic moment: 1- The first one involves an electric and / or circular current having some current density distribution 2- The second origin of magnetic moment creation involves particles having their own angular momentum, the so-called spins. In most cases we are more interested in magnetic dipole moment μμ = N ∙ I ∙ A

where N is the number of turns in the coil, I is the current in the coil and A is the area of the coil. Obviously, the unit of magnetic dipole moment is [A ∙ m2]. Combining the equations μ = N ∙ I ∙ A

and τ = I ∙ B ∙ A ∙ sinθ

we obtain for the magnetic torque in terms of the magnetic dipole moment τ = μ ∙ B ∙ sinθ

The vector form of the above equation is τ*⃗* = μ*⃗* × B*⃗*

A magnetic dipole in an external field has an energy that depends on the orientation of dipole moment in respect to the magnetic field direction. The energy U of magnetic dipoles in terms of the angle θ is: U(θ) = -μ*⃗* ∙ B*⃗*

or U(θ) = -μ ∙ B ∙ cosθ

The sign minus is because the maximum value of energy is obtained for θ = 1800 for which cos θ = -1. In this way, we obtain a positive value for the maximum energy. If an external force rotates the coil (magnetic dipole) from an angle θ1 to another angle θ2, a resulting torque is produced. This action results in a work W done on the dipole by the applied torque. If the dipole is stationary before and after the change in the angle mentioned above, then the work done on the magnetic dipole is W = U_final-U_initial = U(θ_{2} )-U(θ_{1})

where U(θ1) and U(θ2) are the values of energy for the angles θ1 and θ2 formed by the corresponding vectors of magnetic dipole moments and the magnetic field lines. **1)** What is the magnetic dipole moment (magnitude and direction) of the coil shown in the figure?

- 0.8 A ∙ m2 (down)
- 0.008 A ∙ m2 (up)
- 0.025 A ∙ m2 (up)
- 0.001 A ∙ m2 (down)

**Correct Answer: D**

**2)** What is the torque produced on the coil shown in the figure if the coil has 6 turns and 0.25A current is flowing through it? The coil is leaned by 370 to the direction of the 0,75T uniform magnetic field lines and the diameter of coil is 2 cm. (Take cos 370 = 0.8 and sin 370 = 0.6).

- 3.53 × 10-3 N ∙ m
- 3.53 × 10-5 N ∙ m
- 1.13 × 10-5 N ∙ m
- 1.13 × 10-3 N ∙ m

**Correct Answer: B**

**3)** What is the energy possessed by the coil shown in the figure if the diameter of a single loop is 6 cm, the current flowing through the coil is 0.4 A and the (uniform) magnetic field between the two magnets is 4T. The coil is placed horizontally and the magnets vertically.

- 0 J
- -2.939 × 10-2 J
- 2.939 × 10-2 J
- 1.13 × 10-3 J

**Correct Answer: C**

We hope you found this Physics tutorial "Magnetic Dipole Moment" useful. If you thought the guide useful, it would be great if you could spare the time to rate this tutorial and/or share on social media, this helps us identify popular tutorials and calculators and expand our free learning resources to support our users around the world have free access to expand their knowledge of physics and other disciplines. In our next tutorial, we expand your insight and knowledge of Magnetism with our Physics tutorial on Ampere's Law .

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