# 22.8 Torque on a current loop: motors and meters  (Page 2/4)

 Page 2 / 4

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 [link] (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 [link] .)

Meters , such as those in analog fuel gauges on a car, are another common application of magnetic torque on a current-carrying loop. [link] 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 $I$ 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.

## Section summary

• The torque $\tau$ on a current-carrying loop of any shape in a uniform magnetic field. is
$\tau =\text{NIAB}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta ,$
where $N$ is the number of turns, $I$ is the current, $A$ is the area of the loop, $B$ is the magnetic field strength, and $\theta$ is the angle between the perpendicular to the loop and the magnetic field.

## Conceptual questions

Draw a diagram and use RHR-1 to show that the forces on the top and bottom segments of the motor’s current loop in [link] are vertical and produce no torque about the axis of rotation.

## Problems&Exercises

(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) $\text{τ}$ decreases by 5.00% if B decreases by 5.00%

(b) 5.26% increase

(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 $\theta$ is $\text{10}\text{.}9º?$

Find the current through a loop needed to create a maximum torque of $9\text{.}\text{00 N}\cdot \text{m.}$ 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

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

Since the equation for torque on a current-carrying loop is $\tau =\text{NIAB}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta$ , the units of $N\cdot m$ must equal units of $A\cdot {m}^{2}\phantom{\rule{0.25em}{0ex}}T$ . Verify this.

$A\cdot {m}^{2}\cdot T=A\cdot {m}^{2}\left(\frac{N}{A\cdot m}\right)=N\cdot m$ .

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

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 $0\text{.}\text{650}×{\text{10}}^{-\text{15}}\phantom{\rule{0.25em}{0ex}}m$ in radius with a current of $1\text{.}\text{05}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}A$ (no kidding). Find the maximum torque on a proton in a 2.50-T field. (This is a significant torque on a small particle.)

$3\text{.}\text{48}×{\text{10}}^{-\text{26}}\phantom{\rule{0.25em}{0ex}}N\cdot m$

(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. The Earth’s field here is due north, parallel to the ground, with a strength of $3\text{.}\text{00}×{\text{10}}^{-5}\phantom{\rule{0.25em}{0ex}}T$ . What are the direction and magnitude of the torque on the loop? (b) Does this device have any practical applications as a motor?

Repeat [link] , but with the loop lying flat on the ground with its current circulating counterclockwise (when viewed from above) in a location where the Earth’s field is north, but at an angle $\text{45}\text{.}0º$ below the horizontal and with a strength of $\text{6.}\text{00}×{\text{10}}^{-5}\phantom{\rule{0.25em}{0ex}}T$ .

(a) $\text{0.666 N}\cdot m$ west

(b) This is not a very significant torque, so practical use would be limited. Also, the current would need to be alternated to make the loop rotate (otherwise it would oscillate).

#### Questions & Answers

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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