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Physics Lesson 16.5.2 - Torque in Terms of Magnetic Dipole Moment

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Welcome to our Physics lesson on Torque in Terms of Magnetic Dipole Moment, this is the second lesson of our suite of physics lessons covering the topic of Magnetic Dipole Moment, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

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 ∙ sin⁡θ

or

τmax = I ∙ B ∙ A

(this occurs when the angle θ formed by the magnetic field and current is 90°, i.e. when sin θ = 1)

Physics Tutorials: This image provides visual information for the physics tutorial Magnetic Dipole Moment

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

τ⃗ = μ⃗ × B⃗

which is very similar to the torque produced by an electric field E⃗ on an electric dipole moment p⃗

τ⃗ = p⃗ × E⃗

In either case, the torque is the vector product of the specific dipole to the corresponding field.

Example 2

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.

Physics Tutorials: This image provides visual information for the physics tutorial Magnetic Dipole Moment

Calculate:

  1. The magnetic dipole moment produced at the loop
  2. The torque on the loop produced by the magnetic field

Solution 2

  1. 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 ∙ r2
    = 1 ∙ (0.5A) ∙ 3.14 ∙ (0.02m)2
    = 0.000628 A ∙ m2
    = 6.28 × 10-4 A ∙ m2
  2. 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 180°. Hence, we obtain for the torque on the loop:
    τ = μ ∙ B ∙ sin⁡θ
    = 6.28 × 10-4 A ∙ m2 ∙ 0.2T ∙ sin⁡1800
    = 6.28 × 10-4 A ∙ m2 ∙ 0.2T ∙ 0
    = 0
    Physics Tutorials: This image provides visual information for the physics tutorial Magnetic Dipole Moment

This result means no turning effect is produced on the loop.

You have reached the end of Physics lesson 16.5.2 Torque in Terms of Magnetic Dipole Moment. There are 4 lessons in this physics tutorial covering Magnetic Dipole Moment, you can access all the lessons from this tutorial below.

More Magnetic Dipole Moment Lessons and Learning Resources

Magnetism Learning Material
Tutorial IDPhysics Tutorial TitleTutorialVideo
Tutorial		Revision
Notes		Revision
Questions
16.5Magnetic Dipole Moment
Lesson IDPhysics Lesson TitleLessonVideo
Lesson
16.5.1What is Magnetic Moment? Magnetic Dipole Moment
16.5.2Torque in Terms of Magnetic Dipole Moment
16.5.3Energy of a Magnetic Dipole
16.5.4Work Done on a Magnetic Dipole

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