# 22.8 Torque on a current loop: motors and meters

 Page 1 / 4
• Describe how motors and meters work in terms of torque on a current loop.
• Calculate the torque on a current-carrying loop in a magnetic field.

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 [link] .)

Let us examine the force on each segment of the loop in [link] 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. [link] shows views of the loop from above. Torque is defined as $\tau =\text{rF}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta$ , 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 [link] (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 $\left(w/2\right)F\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta$ , and the two add to give a total torque.

$\tau =\frac{w}{2}F\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta +\frac{w}{2}F\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\text{wF}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta$

Now, each vertical segment has a length $l$ that is perpendicular to $B$ , so that the force on each is $F=\text{IlB}$ . Entering $F$ into the expression for torque yields

$\tau =\text{wIlB}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta .$

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 $A=\text{wl}$ ; the expression for the torque becomes

$\tau =\text{NIAB}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta .$

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 $\tau =\text{NIAB}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta$ . Maximum torque occurs when $\theta =\text{90º}$ and $\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =1$ .

Solution

For $\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =1$ , the maximum torque is

${\tau }_{\text{max}}=\text{NIAB}.$

Entering known values yields

$\begin{array}{lll}{\tau }_{\text{max}}& =& \left(\text{100}\right)\left(\text{15.0 A}\right)\left(\text{0.100}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}\right)\left(2\text{.}\text{00 T}\right)\\ & =& \text{30.0 N}\cdot m.\end{array}$

Discussion

This torque is large enough to be useful in a motor.

Pls guys am having problem on these topics: latent heat of fusion, specific heat capacity and the sub topics under them.Pls who can help?
Thanks George,I appreciate.
hamidat
this will lead you rightly of the formula to use
Abolarin
Most especially it is the calculatory aspects that is giving me issue, but with these new strength that you guys have given me,I will put in my best to understand it again.
hamidat
you can bring up a question and let's see what we can do to it
Abolarin
the distance between two suasive crests of water wave traveling of 3.6ms1 is 0.45m calculate the frequency of the wave
v=f×lemda where the velocity is given and lends also given so simply u can calculate the frequency
Abdul
You are right my brother, make frequency the subject of formula and equate the values of velocity and lamda into the equation, that all.
hamidat
lExplain what happens to the energy carried by light that it is dimmed by passing it through two crossed polarizing filters.
When light is reflected at Brewster's angle from a smooth surface, it is 100% polarizedparallel to the surface. Part of the light will be refracted into the surface.
Ekram
What is specific heat capacity?
Specific heat capacity is the amount of heat required to raise the temperature of one (Kg) of a substance through one Kelvin
Paluutar
formula for measuring Joules
I don't understand, do you mean the S.I unit of work and energy?
hamidat
what are the effects of electric current
What limits the Magnification of an optical instrument?
Lithography is 2 micron
Venkateshwarlu
what is expression for energy possessed by water ripple
what is hydrolic press
An hydraulic press is a type of machine that is operated by different pressure of water on pistons.
hamidat
what is dimensional unite of mah
i want jamb related question on this asap🙏
What is Boyles law
it can simple defined as constant temperature
Boyles law states that the volume of a fixed amount of a gas is inversely proportional to the pressure acting on in provided that the temperature is constant.that is V=k(1/p) or V=k/p
what is motion
getting notifications for a dictionary word, smh
Anderson
what is escape velocity
the minimum thrust that an object must have in oder yo escape the gravitational pull
Joshua
what is a dimer
Mua
what is a atom
how to calculate tension
what are the laws of motion
Mua