# 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

Why is there no 2nd harmonic in the classical electron orbit?
Shree Reply
how to reform magnet after been demagneted
Inuwa Reply
A petrol engine has a output of 20 kilowatts and uses 4.5 kg of fuel for each hour of running. The energy given out when 1 kg of petrol is burnt is 4.8 × 10 to the power of 7 Joules. a) What is the energy output of the engine every hour? b) What is the energy input of the engine every hour?
Morris Reply
what is the error during taking work done of a body..
Aliyu Reply
what kind of error do you think? and work is held by which force?
Daniela
I am now in this group
smart
theory,laws,principles and what-a-view are not defined. why? you
Douglas Reply
A simple pendulum is used in a physics laboratory experiment to obtain an experimental value for the gravitational acceleration, g . A student measures the length of the pendulum to be 0.510 meters, displaces it 10 o from the equilibrium position, and releases it. Using a s
Emmanuel Reply
so what question are you passing across... sir?
Olalekan
Two masses of 2 kg and 4 kg are held with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a velocity of 6m/s, what is the stored energy in the spring when it is compressed?
Emmanuel Reply
54 joule
babar
how?
rakesh
Reduce that two body problem into one body problem. Apply potential and k. E formula to get total energy of the system
rakesh
i dont think dere is any potential energy... by d virtue of no height present
Olalekan
there is compressed energy,dats only potential energy na?
rakesh
yes.. but... how will u approach that question without The Height in the question?
Olalekan
Can you explain how you get 54J?
Emmanuel
Because mine is 36J
Emmanuel
got 36J too
Douglas
OK the answer is 54J Babar is correct
Emmanuel
Conservation of Momentum
Emmanuel
woow i see.. can you give the formula for this
joshua
Two masses of 2 kg and 4 kg are held with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a velocity of 6m/s, what is the stored energy in the spring when it is compressed? Asume there is no external force.
Emmanuel Reply
Please help!
Emmanuel
please help find dy/dx 2x-y/x+y
Inuwa
By using the Quotient Rule dy/dx = 3y/(x +y)²
Emmanuel
3y/(x+y)²
Emmanuel
may be by using MC^2=MC^2 and Total energy=kinetic energy +potential energy so 1st find kinetic energy and den find potential energy which is stored energy
rakesh
i think i m correct
rakesh
But how?
Emmanuel
3y/(x+y)²
Douglas
what's the big bang?
kwame Reply
yes what is it?
LamaBbake
it is the explanation of how the universe began
Zainab
yes
Ana
explain
Chinagorom
in
Chinagorom
it is a theory on how the universe began. to understand more I would suggest researching the topic online.
david
thanks guys
kwame
if a force of 12N is applied to load of 200g what us the work done
Joshua Reply
We can seek accelation first
Nancy
we are given f=12 m=200g which is 0.2kg now from 2nd law of newton a= f/m=60m/s*2 work done=force applied x displacement cos (theta) w= 12x60 =720nm/s*2
Mudang
this very interesting question very complicated for me, í need urgent help. 1,two buses A and B travel along the same road in the same direction from Harper city (asume They both started from the same point) to Monrovia. if bus A maintains a Speedy of 60km/h and bus B a Speedy of 75km/h, how many
mohammed
hours Will it take bus B to overtake bus A assuming bus B starts One hour after bus A started. what is the distance travelled by the buses when They meet?.
mohammed
pls í need help
mohammed
4000 work is done
Ana
speed=distance /time distance=speed/time
Ana
now use this formula
Ana
what's the answer then
Julius
great Mudang
Kossi
please Ana explain 4000 ?
babar
hey mudang there is a product of force and acceleration not force and displacement
babar
@Mohammed answer is 0.8hours or 48mins
Douglas
nice
A.d
its not possible
Olalekan
í want the working procedure
mohammed
the answer is given but how Will One arrive at it. the answers are 4hours and 300m.
mohammed
physics is the science that studies the non living nature
isidor Reply
ancient greek language physis = nature
isidor
what is phyacs
technical Reply
if i am going to start studying physics where should i start?
BRIAN Reply
I think from kinematics
Nancy
You can find physics books at the library or online. That's how I started.
Chelsea
And yes, kinematics is usually where you can begin.
Chelsea
study basic algebra and calculus and can start from classical mechanics
Mudang
yes think so but dimension is the best starting point
Obed
3 formula's of equations of motion
benjamin Reply
vf=vi+at........1 s=vit+1/2(at)2 vf2=vi2+2as
Ana
solve the formula's please
benjamin
those are the three .. what you wanna solve ?
Nihrantz
For first equation simply integrate formula of acceleration in the limit v and u
Tripti
For second itegrate velocity formula by ising first equation
Tripti
similarly for 3 one integrate acceleration again by multiplying and dividing term ds
Tripti
any methods can take to solve this eqtions
a=vf-vi/t vf-vi=at vf=vi+at......1
Ana
suppose a body starts with an initial velocity vi and travels with uniform acceleration a for a period of time t.the distance covered by a body in this time is "s" and its final velocity becomes vf
Ana
what is the question dear
Zeeshan
average velocity=(vi+vf)/2 distance travelled=average velocity ×time therefore s=vi+vf/2×t from the first equation of motion ,we have vf =vi+at s=[vi+(vi+at)]/2×t s=(2vi+at)/2×t s=bit+1/2at2
Ana
find the distance
Ana
how
Zeeshan
Two speakers are arranged so that sound waves with the same frequency are produced and radiated through a room. An interference pattern is created. Calculate the distance between the two speakers?
Hayne Reply
How can we calculate without any information?
Amir
I think the formulae used for this question is lambda=(ax)/D
Amir

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