Torque on a Current Loop: Motors and Meters
By the end of this section, you will be able to:
Motors are the most common application of magnetic force on current-carrying wires. Motors have 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 to mechanical work in the process. (See Figure 1)
Let us examine the force on each segment of the loop in Figure 1 to find the torques produced about the axis of the vertical shaft. (This will lead to a useful equation for the torque on the loop.) We take the magnetic field to be uniform over the rectangular loop, which has width w and height \(l\). First, we note that the forces on the top and bottom segments are vertical and, therefore, parallel to the shaft, producing no torque. Those vertical forces are equal in magnitude and opposite in direction, so that they also produce no net force on the loop. Figure 2 shows views of the loop from above. Torque is defined as \(\tau = rF \sin \theta\), where \(F\) is the force, \(r\) is the distance from the pivot that the force is applied, and \(\theta\) is the angle between \(r\) and \(F\). As seen in Figure 2(a), right hand rule 1 gives the forces on the sides to be equal in magnitude and opposite in direction, so that the net force is again zero. However, each force produces a clockwise torque. Since \(r = w/2\), the torque on each vertical segment is \((w/2)F \sin \theta\), and the two add to give a total torque.
Now, each vertical segment has a length \(l\) that is perpendicular to \(B\), so that the force on each is \(F = IlB\). Entering \(F\) into the expression for torque yields
If we have a multiple loop of \(N\) turns, we get \(N\) times the torque of one loop. Finally, note that the area of the loop is \(A = wl\); the expression for the torque becomes
This is the torque on a current-carrying loop in a uniform magnetic field. This equation can be shown to be valid for a loop of any shape. The loop carries a current \(I\), has \(N\) turns, each of area \(A\), and the perpendicular to the loop makes an angle \(\theta\) with the field \(B\). The net force on the loop is zero.
CALCULATING TORQUE ON A CURRENT-CARRYING LOOP IN A STRONG MAGNETIC FIELD
Find the maximum torque on a 100-turn square loop of a wire of 10.0 cm on a side that carries 15.0 A of current in a 2.00-T field.
Strategy
Torque on the loop can be found using \(\tau = NIAB \sin \theta\). Maximum torque occurs when \(\theta = 90^\circ\) and \(\sin \theta =1\).
Solution
For \(\sin \theta =1\), the maximum torque is
Entering known values yields
Discussion
This torque is large enough to be useful in a motor.
The torque found in the preceding example is the maximum. As the coil rotates, the torque decreases to zero at \(\theta = 0\). The torque then reverses its direction once the coil rotates past \(\theta = 0\). (See Figure 2(d).) This means that, unless we do something, the coil will oscillate back and forth about equilibrium at \(\theta = 0\). To get the coil to continue rotating in the same direction, we can reverse the current as it passes through \(\theta = 0\) with automatic switches called brushes. (See Figure 3.)
Meters, such as those in analog fuel gauges on a car, are another common application of magnetic torque on a current-carrying loop. Figure 4 shows that a meter is very similar in construction to a motor. The meter in the figure has its magnets shaped to limit the effect of \(\theta\) by making \(B\) perpendicular to the loop over a large angular range. Thus the torque is proportional to \(l\) and not \(\theta\). A linear spring exerts a counter-torque that balances the current-produced torque. This makes the needle deflection proportional to \(I\). If an exact proportionality cannot be achieved, the gauge reading can be calibrated. To produce a galvanometer for use in analog voltmeters and ammeters that have a low resistance and respond to small currents, we use a large loop area \(A\), high magnetic field \(B\), and low-resistance coils.
Discuss the similarity of the actions of a galvanometer and a motor. Explain your answer.
Which of the following factors affect the magnetic field strength of a coil? Choose 3 correct options.
Calculate the magnetic field strength needed on a 200-turn square loop 20.0 cm on a side to create a maximum torque of \(300\; N\cdot m\) if the loop is carrying 25.0 A.
\(T\)
\(\tau=NIAB\;\;\Rightarrow\;\; B=\frac{\tau}{NIA}\)
\(B=\frac{300\;N\cdot m}{(200)(25.0\; A)(0.200\; m)^2}\)
Answer: 1.50 T
Find the current through a loop needed to create a maximum torque of \(9.00\; N\cdot m\). The loop has 50 square turns that are 15.0 cm on a side and is in a uniform 0.800-T magnetic field.
\(A\)
\(\tau= NIAB\;\; \Rightarrow\;\; I=\frac{\tau}{NAB}\)
\(I=\frac{9.00\;N}{(50)(0.150\;m)^2(0.800\;T)}\)
Answer: 10.0 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?
\(N \cdot m\)
The maximum torque occurs when \(sin \phi= 1\), so the maximum torque is:
\(\tau_{max}=NIABsin\phi \)
\(\tau_{max}=(150)(50.0\;A)(0.180\;m)^2(1.60\; T)(1)\)
Answer: 388.8 \(N \cdot m\)
What is the torque when \(\theta\) is 10.9°?
\(\tau=NIABsin\phi\)
Now set \(\phi=19.9^\circ\), so
\(\tau=(150)(50.0\;A)(0.180\; m)^2(1.60\; T)(sin(10.9^{\circ})\)
Answer: 73.52 \(N\cdot m\)
At what angle \(\theta\) is the torque on a current loop 90.0% of the maximum?
\(^\circ\)
\(\theta=sin^{-1}(0.90)\)
Answer: 64.1580°
At what angle \(\theta\) is the torque on a current loop 50.0% of the maximum?
\(\theta= sin^{-1}(0.50)\)
Answer: 30.0°
At what angle \(\theta\) is the torque on a current loop 10.0% of the maximum?
\(\theta= sin^{-1}(0.10)\)
Answer: 5.7391°
A proton has a magnetic field due to its spin on its axis. The field is similar to that created by a circular current loop \(0.650\times10^{-15}\;m\) in radius with a current of \(1.05\times10^4\;A\) (no kidding). Find the maximum torque on a proton in a 2.50-T field. (This is a significant torque on a small particle.)
\(\tau=IAB\)
\(\tau=(1.05\times10^4\;A)(\pi)(0.650\times10^{-15}\;m)^2(2.50\; T)\)
Answer: 3.48E-26 \(N\cdot m\)
Magnetic Fields Produced by Currents: Ampere’s Law
How much current is needed to produce a significant magnetic field, perhaps as strong as the Earth’s field? Surveyors will tell you that overhead electric power lines create magnetic fields that interfere with their compass readings. Indeed, when Oersted discovered in 1820 that a current in a wire affected a compass needle, he was not dealing with extremely large currents. How does the shape of wires carrying current affect the shape of the magnetic field created? We noted earlier that a current loop created a magnetic field similar to that of a bar magnet, but what about a straight wire or a toroid (doughnut)? How is the direction of a current-created field related to the direction of the current? Answers to these questions are explored in this section, together with a brief discussion of the law governing the fields created by currents.
Magnetic fields have both direction and magnitude. As noted before, one way to explore the direction of a magnetic field is with compasses, as shown for a long straight current-carrying wire in Figure 1. Hall probes can determine the magnitude of the field. The field around a long straight wire is found to be in circular loops. The right hand rule 2 (RHR-2) emerges from this exploration and is valid for any current segment—point the thumb in the direction of the current, and the fingers curl in the direction of the magnetic field loops created by it.
For a wire oriented perpendicular to the page, if the current in the wire is directed out of the page, the right-hand rule tells us that the magnetic field lines will be oriented in a counterclockwise direction around the wire. If the current in the wire is directed into the page, the magnetic field lines will be oriented in a clockwise direction around the wire. We use ⊙ to indicate that the direction of the current in the wire is out of the page, and ⊗ for the direction into the page.
The magnetic field strength (magnitude) produced by a long straight current-carrying wire is found by experiment to be
where \(l\) is the current, \(r\) is the shortest distance to the wire, and the constant \(\mu _{0} = 4 \pi \times 10^{-7} \thinspace \mathrm{T \cdot m/A}\) is the permeability of free space. (\(\mu _{0}\) is one of the basic constants in nature. We will see later that \(\mu _{0}\) is related to the speed of light.) Since the wire is very long, the magnitude of the field depends only on distance from the wire \(r\), not on position along the wire.
CALCULATING CURRENT THAT PRODUCES A MAGNETIC FIELD
Find the current in a long straight wire that would produce a magnetic field twice the strength of the Earth’s at a distance of 5.0 cm from the wire.
The Earth’s field is about \(5.0 \times 10^{-5} \mathrm{T}\), and so here \(B\) due to the wire is taken to be \(1.0 \times 10^{-4} \mathrm{T}\). The equation \(B = \frac{\mu _{0}I}{2 \pi r}\) can be used to find \(I\), since all other quantities are known.
Solving for \(I\) and entering known values gives
So a moderately large current produces a significant magnetic field at a distance of 5.0 cm from a long straight wire. Note that the answer is stated to only two digits, since the Earth’s field is specified to only two digits in this example.
The magnetic field of a long straight wire has more implications than you might at first suspect. Each segment of current produces a magnetic field like that of a long straight wire, and the total field of any shape current is the vector sum of the fields due to each segment. The formal statement of the direction and magnitude of the field due to each segment is called the Biot-Savart law. Integral calculus is needed to sum the field for an arbitrary shape current. This results in a more complete law, called Ampere’s law, which relates magnetic field and current in a general way. Ampere’s law in turn is a part of Maxwell’s equations, which give a complete theory of all electromagnetic phenomena. Considerations of how Maxwell’s equations appear to different observers led to the modern theory of relativity, and the realization that electric and magnetic fields are different manifestations of the same thing. Most of this is beyond the scope of this text in both mathematical level, requiring calculus, and in the amount of space that can be devoted to it. But for the interested student, and particularly for those who continue in physics, engineering, or similar pursuits, delving into these matters further will reveal descriptions of nature that are elegant as well as profound. In this text, we shall keep the general features in mind, such as RHR-2 and the rules for magnetic field lines listed in Magnetic Fields and Magnetic Field Lines, while concentrating on the fields created in certain important situations.
Hearing all we do about Einstein, we sometimes get the impression that he invented relativity out of nothing. On the contrary, one of Einstein’s motivations was to solve difficulties in knowing how different observers see magnetic and electric fields.
The magnetic field near a current-carrying loop of wire is shown in Figure 3. Both the direction and the magnitude of the magnetic field produced by a current-carrying loop are complex. RHR-2 can be used to give the direction of the field near the loop, but mapping with compasses and the rules about field lines given in Magnetic Fields and Magnetic Field Lines are needed for more detail. There is a simple formula for the magnetic field strength at the center of a circular loop. It is
where \(R\) is the radius of the loop. This equation is very similar to that for a straight wire, but it is valid only at the center of a circular loop of wire. The similarity of the equations does indicate that similar field strength can be obtained at the center of a loop. One way to get a larger field is to have \(N\) loops; then, the field is \(B = N_{\mu_{0}} I /(2R)\). Note that the larger the loop, the smaller the field at its center, because the current is farther away.
A solenoid is a long coil of wire (with many turns or loops, as opposed to a flat loop). Because of its shape, the field inside a solenoid can be very uniform, and also very strong. The field just outside the coils is nearly zero. Figure 4 shows how the field looks and how its direction is given by RHR-2.
The magnetic field inside of a current-carrying solenoid is very uniform in direction and magnitude. Only near the ends does it begin to weaken and change direction. The field outside has similar complexities to flat loops and bar magnets, but the magnetic field strength inside a solenoid is simply
where \(n\) is the number of loops per unit length of the solenoid (\(n = N/l\), with \(N\) being the number of loops and \(l\) the length). Note that \(B\) is the field strength anywhere in the uniform region of the interior and not just at the center. Large uniform fields spread over a large volume are possible with solenoids, as Example #2 implies.
CALCULATING FIELD STRENGTH INSIDE A SOLENOID
What is the field inside a 2.00-m-long solenoid that has 2000 loops and carries a 1600-A current?
To find the field strength inside a solenoid, we use \(B = \mu _{0}nI\). First, we note the number of loops per unit length is
Substituting known values gives
This is a large field strength that could be established over a large-diameter solenoid, such as in medical uses of magnetic resonance imaging (MRI). The very large current is an indication that the fields of this strength are not easily achieved, however. Such a large current through 1000 loops squeezed into a meter’s length would produce significant heating. Higher currents can be achieved by using superconducting wires, although this is expensive. There is an upper limit to the current, since the superconducting state is disrupted by very large magnetic fields.
Visit here and start the simulation applet “Particle in a Magnetic Field (2D)” in order to explore the magnetic force that acts on a charged particle in a magnetic field. Experiment with the simulation to see how it works and what parameters you can change; then construct a plan to methodically investigate how magnetic fields affect charged particles. Some questions you may want to answer as part of your experiment are:
There are interesting variations of the flat coil and solenoid. For example, the toroidal coil used to confine the reactive particles in tokamaks is much like a solenoid bent into a circle. The field inside a toroid is very strong but circular. Charged particles travel in circles, following the field lines, and collide with one another, perhaps inducing fusion. But the charged particles do not cross field lines and escape the toroid. A whole range of coil shapes are used to produce all sorts of magnetic field shapes. Adding ferromagnetic materials produces greater field strengths and can have a significant effect on the shape of the field. Ferromagnetic materials tend to trap magnetic fields (the field lines bend into the ferromagnetic material, leaving weaker fields outside it) and are used as shields for devices that are adversely affected by magnetic fields, including the Earth’s magnetic field.
Generate electricity with a bar magnet! Discover the physics behind the phenomena by exploring magnets and how you can use them to make a bulb light.
Click here to view the simulation page
Make a drawing in your notebook and use RHR-2 to find the direction of the magnetic field of a current loop in a motor (such as in the figure below). Then show that the direction of the torque on the loop is the same as produced by like poles repelling and unlike poles attracting.
Figure 1. Torque on a current loop. A current-carrying loop of wire attached to a vertically rotating shaft feels magnetic forces that produce a clockwise torque as viewed from above.