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Physics Lesson 16.3.4 - Magnetic Force on a Current Carrying Loop. The Motor Effect

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Welcome to our Physics lesson on Magnetic Force on a Current Carrying Loop. The Motor Effect, this is the fourth lesson of our suite of physics lessons covering the topic of Magnetic Force on a Current Carrying Wire. Ampere's Force, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

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Magnetic Force on a Current Carrying Loop. The Motor Effect

Let's consider a rectangular loop placed between the opposite poles of two magnets as shown in the figure.

Physics Tutorials: This image provides visual information for the physics tutorial Magnetic Force on a Current Carrying Wire. Ampere's Force

A current I flows through the loop as shown by the yellow arrows. The magnetic field lines lie from North to South poles of magnets (from right to left). Using the Fleming's Left Hand Rule (magnetic field lines punch the palm, the four fingers show the current, the thumb shows the magnetic force), we see that a torque is produced in the two lateral sides of the loop because current flows in opposite directions.

To calculate the torque on the loop we must calculate the forces acting at the lateral sides of the loop only because the front and the back sides are parallel to the magnetic field lines and therefore, they don't produce any force (sin 00 = 0, so F = I ∙ B ∙ L ∙ sin 00 = 0). Let's denote the four vertices of the loop by a, b, c and d respectively. The turning arm is therefore 'ad'/2 or 'bc'/2. If we denote by L the loop sides ab and bc in which the magnetic force produces rotation, we obtain for the scalar version of magnetic forces F1 and F2:

F1 = I ∙ B ∙ ('ab') ∙ sin⁡900
= I ∙ B ∙ L

and

F2 = I ∙ B ∙ ('cd') ∙ sin⁡900
= I ∙ B ∙ L

Therefore, the magnetic forces F1 and F2 are equal in magnitude and opposite in direction. If we denote by x/2 the distance from the lateral sides ab and bc to the rotating axis passing through the centre of loop, we obtain for the maximum torque τ produced:

τmax = F1 ∙ x/2 + F2 ∙ x/2
= I ∙ B ∙ L ∙ x/2 + I ∙ B ∙ L ∙ x/2
= 2I ∙ B ∙ L ∙ x/2
= I ∙ B ∙ L ∙ x

where x is the width of the loop (it represents the side lengths ad or bc).

Since L ∙ x gives the area A of the loop, we can write the last equation as

τmax = I ∙ B ∙ A

Example 4

What is the maximum torque produced by the rectangular loop placed between the opposite poles of two magnets shown in the figure? The magnetic field between the opposite poles is 5mT and the current flowing through the loop is 4A. The dimensions of the loop are 30cm × 20cm. What is the turning direction of the loop?

Physics Tutorials: This image provides visual information for the physics tutorial Magnetic Force on a Current Carrying Wire. Ampere's Force

Solution 4

Clues:

I = 4A
B = 5mT = 5 × 10-3 T
A = 30 cm × 20 cm = 600 cm2 = 0.06 m2 = 6 × 10-2 m2
τmax = ?

The maximum torque produced by this system is

τmax = I ∙ B ∙ A
= (4A) ∙ (5 × 10-3 T) ∙ (6 × 10-2 m2 )
= 120 × 10-5 N ∙ m
= 0.0012 N ∙ m

You have reached the end of Physics lesson 16.3.4 Magnetic Force on a Current Carrying Loop. The Motor Effect. There are 5 lessons in this physics tutorial covering Magnetic Force on a Current Carrying Wire. Ampere's Force, you can access all the lessons from this tutorial below.

More Magnetic Force on a Current Carrying Wire. Ampere's Force Lessons and Learning Resources

Magnetism Learning Material
Tutorial IDPhysics Tutorial TitleTutorialVideo
Tutorial		Revision
Notes		Revision
Questions
16.3Magnetic Force on a Current Carrying Wire. Ampere's Force
Lesson IDPhysics Lesson TitleLessonVideo
Lesson
16.3.1Magnetic Force on a Current Carrying Wire. Ampere's Force
16.3.2The Direction of Magnetic Force. Fleming's Left Hand Rule
16.3.3Magnetic Force between two Parallel Current Carrying Conducting Wires
16.3.4Magnetic Force on a Current Carrying Loop. The Motor Effect
16.3.5Electric Motor

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