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The torque found in the preceding example is the maximum. As the coil rotates, the torque decreases to zero at . The torque then reverses its direction once the coil rotates past . (See [link] (d).) This means that, unless we do something, the coil will oscillate back and forth about equilibrium at . To get the coil to continue rotating in the same direction, we can reverse the current as it passes through 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 by making perpendicular to the loop over a large angular range. Thus the torque is proportional to and not . A linear spring exerts a counter-torque that balances the current-produced torque. This makes the needle deflection proportional to . 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 , high magnetic field , and low-resistance coils.
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.
(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) 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 is
Find the current through a loop needed to create a maximum torque of 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 if the loop is carrying 25.0 A.
Since the equation for torque on a current-carrying loop is , the units of must equal units of . Verify this.
.
(a) At what angle 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 in radius with a current of (no kidding). Find the maximum torque on a proton in a 2.50-T field. (This is a significant torque on a small particle.)
(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 . 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 below the horizontal and with a strength of .
(a) 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).
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