<< Chapter < Page Chapter >> Page >
  • 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] .)

Diagram showing a current-carrying loop of width w and length l between the north and south poles of a magnet. The north pole is to the left and the south pole is to the right of the loop. The magnetic field B runs from the north pole across the loop to the south pole. The loop is shown at an instant, while rotating clockwise. The current runs up the left side of the loop, across the top, and down the right side. There is a force F oriented into the page on the left side of the loop and a force F oriented out of the page on the right side of the loop. The torque on the loop is clockwise as viewed from above.
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.

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 τ = rF sin θ size 12{τ= ital "rF""sin"θ} {} , where F size 12{F} {} is the force, r is the distance from the pivot that the force is applied, and θ 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 ( w / 2 ) F sin θ , and the two add to give a total torque.

τ = w 2 F sin θ + w 2 F sin θ = wF sin θ size 12{τ= { {w} over {2} } F"sin"θ+ { {w} over {2} } F"sin"θ= ital "wF""sin"θ} {}
Diagram showing a current-carrying loop from the top, and four different times as it rotates in a magnetic field. The magnetic field oriented toward the right, perpendicular to the vertical dimension of the loop. In figure a, the top view of the loop is oriented at an angle to the magnetic field lines, which run left to right. The force on the loop is up on the lower left side where the current comes out of the page. The force is down on the upper right side where the loop goes into the page. The angle between the force and the loop is theta. Torque is clockwise and equals w over 2 times I l B sine theta. Figure b shows the top view of the loop parallel to the magnetic field lines. The force on the loop is up on the left side where I comes out of the page. The force on the loop is down on the right side where I goes into the page. The angle theta between the F and B is ninety degrees. Torque is clockwise and equals w over 2 I l B equals maximum torque. Figure c shows the top view of the loop oriented perpendicular to B. The force on the loop is up at the top, where I comes out of the page, and down at the bottom where I goes into the page. Theta equals 0 degrees. Torque equals zero since sine theta equals 0. In figure d the force is down on the lower left side of the loop where I goes in, and up on the upper right side of the loop where I comes out. The torque is counterclockwise. Torque is negative.
Top views of a current-carrying loop in a magnetic field. (a) The equation for torque is derived using this view. Note that the perpendicular to the loop makes an angle θ size 12{θ} {} with the field that is the same as the angle between w / 2 size 12{w/2} {} and F size 12{F} {} . (b) The maximum torque occurs when θ size 12{θ} {} is a right angle and sin θ = 1 size 12{"sin"θ=1} {} . (c) Zero (minimum) torque occurs when θ size 12{θ} {} is zero and sin θ = 0 . (d) The torque reverses once the loop rotates past θ = 0 .

Now, each vertical segment has a length l size 12{l} {} that is perpendicular to B size 12{B} {} , so that the force on each is F = IlB size 12{F= ital "IlB"} {} . Entering F size 12{F} {} into the expression for torque yields

τ = wIlB sin θ . size 12{τ= ital "wIlB""sin"θ} {}

If we have a multiple loop of N size 12{N} {} turns, we get N size 12{N} {} times the torque of one loop. Finally, note that the area of the loop is A = wl size 12{A= ital "wl"} {} ; the expression for the torque becomes

τ = NIAB sin θ . size 12{τ= ital "NIAB""sin"θ} {}

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 size 12{I} {} , has N size 12{N} {} turns, each of area A size 12{A} {} , and the perpendicular to the loop makes an angle θ size 12{θ} {} with the field B size 12{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 τ = NIAB sin θ size 12{τ= ital "NIAB""sin"θ} {} . Maximum torque occurs when θ = 90º and sin θ = 1 size 12{"sin"θ=1} {} .

Solution

For sin θ = 1 size 12{"sin"θ=1} {} , the maximum torque is

τ max = NIAB . size 12{τ rSub { size 8{"max"} } = ital "NIAB"} {}

Entering known values yields

τ max = 100 15.0 A 0.100 m 2 2 . 00 T = 30.0 N m. alignl { stack { size 12{τ rSub { size 8{"max"} } = left ("100" right ) left ("15" "." 0" A" right ) left (0 "." "100"" m" rSup { size 8{2} } right ) left (2 "." "00"" T" right )} {} #" "="30" "." "0 N" cdot m "." {} } } {}

Discussion

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

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
tijani
what is titration
John Reply
what is physics
Siyaka Reply
A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
Jude Reply
Can you compute that for me. Ty
Jude
what is the dimension formula of energy?
David Reply
what is viscosity?
David
what is inorganic
emma Reply
what is chemistry
Youesf Reply
what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
Adjei
please, I'm a physics student and I need help in physics
Adjanou
chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.
Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
Sahid Reply
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
Ryan
what's motion
Maurice Reply
what are the types of wave
Maurice
answer
Magreth
progressive wave
Magreth
hello friend how are you
Muhammad Reply
fine, how about you?
Mohammed
hi
Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
yasuo Reply
Who can show me the full solution in this problem?
Reofrir Reply
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply
Practice Key Terms 2

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, General physics ii phy2202ca. OpenStax CNX. Jul 05, 2013 Download for free at http://legacy.cnx.org/content/col11538/1.2
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'General physics ii phy2202ca' conversation and receive update notifications?

Ask