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To find the net torque on the current loop shown in [link] , we first consider F 1 and F 3 . Since they have the same line of action and are equal and opposite, the sum of their torques about any axis is zero (see Fixed-Axis Rotation ). Thus, if there is any torque on the loop, it must be furnished by F 2 and F 4 . Let’s calculate the torques around the axis that passes through point O of [link] (a side view of the coil) and is perpendicular to the plane of the page. The point O is a distance x from side 2 and a distance ( a x ) from side 4 of the loop. The moment arms of F 2 and F 4 are x sin θ and ( a x ) sin θ , respectively, so the net torque on the loop is

τ = τ 1 + τ 2 + τ 3 + τ 4 = F 2 x sin θ i ^ F 4 ( a x ) sin ( θ ) i ^ = I b B x sin θ i ^ I b B ( a x ) sin θ i ^ .

This simplifies to

τ = I A B sin θ i ^

where A = a b is the area of the loop.

Notice that this torque is independent of x ; it is therefore independent of where point O is located in the plane of the current loop. Consequently, the loop experiences the same torque from the magnetic field about any axis in the plane of the loop and parallel to the x -axis.

A closed-current loop is commonly referred to as a magnetic dipole    and the term IA is known as its magnetic dipole moment     μ . Actually, the magnetic dipole moment is a vector that is defined as

μ = I A n ^

where n ^ is a unit vector directed perpendicular to the plane of the loop (see [link] ). The direction of n ^ is obtained with the RHR-2—if you curl the fingers of your right hand in the direction of current flow in the loop, then your thumb points along n ^ . If the loop contains N turns of wire, then its magnetic dipole moment is given by

μ = N I A n ^ .

In terms of the magnetic dipole moment, the torque on a current loop due to a uniform magnetic field can be written simply as

τ = μ × B .

This equation holds for a current loop in a two-dimensional plane of arbitrary shape.

Using a calculation analogous to that found in Capacitance for an electric dipole, the potential energy of a magnetic dipole is

U = μ · B .

Forces and torques on current-carrying loops

A circular current loop of radius 2.0 cm carries a current of 2.0 mA. (a) What is the magnitude of its magnetic dipole moment? (b) If the dipole is oriented at 30 degrees to a uniform magnetic field of magnitude 0.50 T, what is the magnitude of the torque it experiences and what is its potential energy?

Strategy

The dipole moment is defined by the current times the area of the loop. The area of the loop can be calculated from the area of the circle. The torque on the loop and potential energy are calculated from identifying the magnetic moment, magnetic field, and angle oriented in the field.

Solution

  1. The magnetic moment μ is calculated by the current times the area of the loop or π r 2 .
    μ = I A = ( 2.0 × 10 −3 A)( π (0.02 m) 2 ) = 2.5 × 10 −6 A · m 2
  2. The torque and potential energy are calculated by identifying the magnetic moment, magnetic field, and the angle between these two vectors. The calculations of these quantities are:
    τ = μ × B = μ B sin θ = ( 2.5 × 10 −6 A · m 2 ) ( 0.50 T ) sin ( 30 ° ) = 6.3 × 10 −7 N · m U = μ · B = μ B cos θ = ( 2.5 × 10 −6 A · m 2 ) ( 0.50 T ) cos ( 30 ° ) = −1.1 × 10 −6 J.

Significance

The concept of magnetic moment at the atomic level is discussed in the next chapter. The concept of aligning the magnetic moment with the magnetic field is the functionality of devices like magnetic motors, whereby switching the external magnetic field results in a constant spinning of the loop as it tries to align with the field to minimize its potential energy.

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Check Your Understanding

In what orientation would a magnetic dipole have to be to produce (a) a maximum torque in a magnetic field? (b) A maximum energy of the dipole?

a. aligned or anti-aligned; b. perpendicular

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Summary

  • The net force on a current-carrying loop of any plane shape in a uniform magnetic field is zero.
  • The net torque τ on a current-carrying loop of any shape in a uniform magnetic field is calculated using τ = μ × B where μ is the magnetic dipole moment and B is the magnetic field strength.
  • The magnetic dipole moment μ is the product of the number of turns of wire N , the current in the loop I , and the area of the loop A or μ = N I A n ^ .

Problems

(a) By how many percent is the torque of a motor decreased if its permanent magnets lose 5.0% of their strength? (b) How many percent would the current need to be increased to return the torque to original values?

a. τ = N I A B , so τ decreases by 5.00% if B decreases by 5.00%; b. 5.26% increase

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(a) What is the maximum torque on a 150-turn square loop of wire 18.0 cm on a side that carries a 50.0-A current in a 1.60-T field? (b) What is the torque when θ is 10.9º?

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Find the current through a loop needed to create a maximum torque of 9.0 N · m . The loop has 50 square turns that are 15.0 cm on a side and is in a uniform 0.800-T magnetic field.

10.0 A

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Calculate the magnetic field strength needed on a 200-turn square loop 20.0 cm on a side to create a maximum torque of 300 N ⋅ m if the loop is carrying 25.0 A.

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Since the equation for torque on a current-carrying loop is τ = NIAB sin θ , the units of N ⋅ m must equal units of A ⋅ m 2 T. Verify this.

A · m 2 · T = A · m 2 . N A · m = N · m

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(a) At what angle θ is the torque on a current loop 90.0% of maximum? (b) 50.0% of maximum? (c) 10.0% of maximum?

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A proton has a magnetic field due to its spin. The field is similar to that created by a circular current loop 0.65 × 10 −15 m in radius with a current of 1.05 × 10 4 A . Find the maximum torque on a proton in a 2.50-T field. (This is a significant torque on a small particle.)

3.48 × 10 −26 N · m

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(a) A 200-turn circular loop of radius 50.0 cm is vertical, with its axis on an east-west line. A current of 100 A circulates clockwise in the loop when viewed from the east. Earth’s field here is due north, parallel to the ground, with a strength of 3.0 × 10 −5 T . What are the direction and magnitude of the torque on the loop? (b) Does this device have any practical applications as a motor?

A vertical circular loop is shown along with the compass directions. The axis (perpendicular to the plane of the loop) is on the east-west line. The magnetic field points north. The current in the loop is circulates clockwise as viewed from the east. The torque on the loop is clockwise as viewed when looking down at the top of the loop.
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Repeat the previous problem, but with the loop lying flat on the ground with its current circulating counterclockwise (when viewed from above) in a location where Earth’s field is north, but at an angle 45.0° below the horizontal and with a strength of 6.0 × 10 −5 T .

A horizontal circular loop is shown along with the compass directions. The magnetic field points 45 degrees below the horizontal. The loop is shown rotating clockwise as viewed from the east.

0.666 N · m

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Practice Key Terms 3

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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