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11.5: Force and Torque on a Current Loop

Motors are the most common application of magnetic force on current-carrying wires. Motors contain loops of wire in a magnetic field. When current is passed through the loops, the magnetic field exerts torque on the loops, which rotates a shaft. Electrical energy is converted into mechanical work in the process. Once the loop’s surface area is aligned with the magnetic field, the direction of current is reversed, so there is a continual torque on the loop (Figure \(\PageIndex{1}\)). This reversal of the current is done with commutators and brushes. The commutator is set to reverse the current flow at set points to keep continual motion in the motor. A basic commutator has three contact areas to avoid and dead spots where the loop would have zero instantaneous torque at that point. The brushes press against the commutator, creating electrical contact between parts of the commutator during the spinning motion.

A schematic of a d c motor consisting of a magnet with a horizontal gap, a power supply with leads attached to brushes, an a wire is bent into a rectangular loop. The ends of the wire are attached to contacts that connect to the brushes of the power supply when the loop is horizontal. When the loop is vertical, they align with the gap between the contacts. The north pole of the magnet is on the left, the south pole on the right. Figure a: The loop is horizontal and the brushes make contact with the loop. A clockwise (looking down) current flows through the loop, so the current in the left segment of the loop flows into the page, and the current in the right segment flows out of the page. The magnetic force on the left segment is down, and on the right segment is up. The loop rotates counterclockwise (looking into the page.) Figure b: The loop is vertical. The brushes are not in contact with the loop. No current flows and no forces are exerted.

Figure \(\PageIndex{1}\): A simplified version of a dc electric motor. (a) The rectangular wire loop is placed in a magnetic field. The forces on the wires closest to the magnetic poles (N and S) are opposite in direction as determined by the right-hand rule-1. Therefore, the loop has a net torque and rotates to the position shown in (b). (b) The brushes now touch the commutator segments so that no current flows through the loop. No torque acts on the loop, but the loop continues to spin from the initial velocity given to it in part (a). By the time the loop flips over, current flows through the wires again but now in the opposite direction, and the process repeats as in part (a). This causes continual rotation of the loop.

In a uniform magnetic field, a current-carrying loop of wire, such as a loop in a motor, experiences both forces and torques on the loop. Figure \(\PageIndex{1}\) shows a rectangular loop of wire that carries a current I and has sides of lengths a and b. The loop is in a uniform magnetic field: \(\vec{B} = B\hat{j}\). The magnetic force on a straight current-carrying wire of length l is given by \(I\vec{l} \times \vec{B}\). To find the net force on the loop, we have to apply this equation to each of the four sides. The force on side 1 is

\[\vec{F}_1 = IaB \sin(90^o - \theta) \hat{i} = IaB \cos\theta \hat{i}\]

where the direction has been determined with the RHR-1. The current in side 3 flows in the opposite direction to that of side 1, so

\[\vec{F}_3 = -IaB \sin(90^o + \theta)\hat{i} = -IaB \cos \theta \hat{i}\]

The currents in sides 2 and 4 are perpendicular to \(\vec{B}\) and the forces on these sides are

\[\vec{F}_2 = IbB\hat{k}\]

\[\vec{F}_4 = -IbB\hat{k}.\]

We can now find the net force on the loop:

\[\sum \vec{F}_{net} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \vec{F}_4 = 0.\]

Although this result \((\sum F = 0)\) has been obtained for a rectangular loop, it is far more general and holds for current-carrying loops of arbitrary shapes; that is, there is no net force on a current loop in a uniform magnetic field.

An illustration of a rectangular loop carrying a current I. The current in the loop is counterclockwise when viewed from the positive y direction looking toward the origin. The loop is in a uniform magnetic field, B, that is pointing to the right. Figure a shows a 3 dimensional view of the loop. The top and bottom sides are parallel to the x axis and have length b. The top side is at y=0 and positive z with current in the positive x direction. The bottom side is at a positive y and z=0 and has current in the negative x direction. The remaining two sides have length b. One is at x=0 and has current going up, and one is at positive x and has current going up. These sides are tilted at an angle theta at the top with respect to the z axis. The direction of the unit vector n hat normal to the area of the rectangular loop is shown. The forces on each of the sides are also shown. F 1 is the force on the tilted side at positive x, and points in the positive x direction. F 2 is the force on the top side and points up. F 3 is the force on the tilted side at x=0 and points in the negative x direction. F 4 is the force on the bottom and points down. Figure b shows a side view of the loop, so that we are looking at the y z plane and see only the tilted side, which makes an angle of theta with the vertical at the top. The current is coming out at us at the top of the loop, and the current is going into the page at the bottom. The force F 2 on the top is up, the force F 4 on the bottom is down. The n hat vector points up and to the right, at an angle of theta to the field B. The pivot point O about which we are calculating the torque is shown a distance x from the top of the loop, and a-x from the bottom.

Figure \(\PageIndex{2}\): (a) A rectangular current loop in a uniform magnetic field is subjected to a net torque but not a net force. (b) A side view of the coil.

To find the net torque on the current loop shown in Figure \(\PageIndex{2a}\), 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 Figure \(\PageIndex{2b}\) (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 \(x \space sin \space \theta\) and \((a - x)\space sin \space \theta\), respectively, so the net torque on the loop is

\[\sum \vec{\tau} = \vec{\tau}_1 + \vec{\tau}_2 + \vec{\tau}_3 + \vec{\tau}_4 = F_2 x \space sin \space \theta \hat{i} - F_4 (a - x) \space sin \space (\theta) \hat{i}\] \[- IbBx \space sin \space \theta \hat{i} - IbB(a - x) sin \space \theta \hat{i}.\]

This simplifies to \[\vec{\tau} = - IAB \space sin \space \theta \hat{i}\] where \(A = ab\) 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 \(\mu\). Actually, the magnetic dipole moment is a vector that is defined as

\[\vec{\mu} = IA \hat{n}\] where \(\hat{n}\) is a unit vector directed perpendicular to the plane of the loop (see Figure). The direction of \(\hat{n}\) 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 \(\hat{n}\). If the loop contains N turns of wire, then its magnetic dipole moment is given by

\[\vec{\mu} = NIA\hat{n}.\]

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

\[\vec{\tau} = \vec{\mu} \times \vec{B}.\]

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

\[U = -\vec{\mu} \cdot \vec{B}.\]

Example \(\PageIndex{1}\): 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 \(\pi r^2\). \[\mu = IA = (2.0 \times 10^{-3} A)(\pi (0.02 \space m)^2) = 2.5 \times 10^{-6} A \cdot m^2\]
  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: \[\tau = \vec{\mu} \times \vec{B} = \mu B \space sin \space \theta = (2.5 \times 10^{-6} A \cdot m^2)(0.50 T) sin(30^o) = 6.3 \times 10^{-7}N \cdot m\] \[U = -\vec{\mu} \cdot \vec{B} = - \mu B cos \theta = - (2.5 \times 10^{-6} A \cdot m^2)(0.50 T) cos (30^o) = -1.1 \times 10^{-6}J.\]

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.

Exercise \(\PageIndex{1}\)

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?

Solution

a. aligned or anti-aligned; b. perpendicular

Contributors

Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).