Magnetic field with loop

  • #1
annamal
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Homework Statement:
Previous problem:
(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. 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 the previous problem, but with the loop lying flat on the ground with its current circulating counterclockwise (when viewed from above) in a location where Earth’s field is north, but at an angle 45.0° below the horizontal and with a strength of ##6*10^-5## T. See image.
Relevant Equations:
##\tau = -IABsin\theta##
I am confused about this, do the black arrows represent the direction of magnetic force?
The torque ##\tau = -IABsin\theta##, where I = current A is area of loop and B is magnetic field strength and I am a little confused how ##\theta## here is 45 degrees when the angle between the normal for the loop and the magnetic field B is 90 degrees unless I am reading the problem incorrectly?

Screen Shot 2022-05-11 at 12.47.46 AM.png
 

Answers and Replies

  • #2
haruspex
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the angle between the normal for the loop and the magnetic field B is 90 degrees
The normal to the loop is vertical and the field is at 45° below the horizontal.
 
  • #3
annamal
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The normal to the loop is vertical and the field is at 45° below the horizontal.
Is 45 degrees below the horizontal supposed to be pointing into the page because I thought the 45 degrees was to the right of the north line. And the normal of the loop seems to be out of the page if we curl our fingers in the direction of the current with the normal being the direction our thumb points.
 
  • #4
haruspex
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Is 45 degrees below the horizontal supposed to be pointing into the page because I thought the 45 degrees was to the right of the north line. And the normal of the loop seems to be out of the page if we curl our fingers in the direction of the current with the normal being the direction our thumb points.
I'm not sure what plane you are calling the "page".
The normal to the loop, as a vector representing the current, is straight up.
A field line through the centre of the loop lies in a vertical plane aligned NS. It angles down into the ground at 45°.
 
  • #5
annamal
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I'm not sure what plane you are calling the "page".
The normal to the loop, as a vector representing the current, is straight up.
A field line through the centre of the loop lies in a vertical plane aligned NS. It angles down into the ground at 45°.
In my image, the loop lies in the plane of N to S and E to W. That is the plane of the page. Into the page would be into and out of the page would be sticking out of the page. See the black arrow below as a representation of the normal. This normal is not at a 45 degree angle to the magnetic field but yet the theta is my torque equation is 45 degrees.

Screen Shot 2022-05-11 at 3.01.32 PM.png

Also, are the black arrows I drew in the original post correct for the magnetic force?
 
  • #6
kuruman
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In my image, the loop lies in the plane of N to S and E to W. That is the plane of the page. Into the page would be into and out of the page would be sticking out of the page. See the black arrow below as a representation of the normal. This normal is not at a 45 degree angle to the magnetic field but yet the theta is my torque equation is 45 degrees.

View attachment 301362
Also, are the black arrows I drew in the original post correct for the magnetic force?
Please confirm that the drawing you have in post #5 corresponds to the last part of the problem that starts with "Repeat the previous problem ##\dots##". If that is the case, then what does the arrow encircling the E-axis signify?

One more question: The normal to the loop forms an angle of 90° with the N-axis and the N-axis forms and angle of 45° with the field. What is the angle between the field and the loop normal? What is the sine of that angle and how does it differ from the sine of 45°?

And one observation: People connected to the internet use electronic devices with flat screens, for the most part, not "pages". Since figures are two dimensional projections onto such screens, it is more appropriate to talk about a direction perpendicular to the screen and either towards or away from the viewer.
 
  • #7
annamal
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Please confirm that the drawing you have in post #5 corresponds to the last part of the problem that starts with "Repeat the previous problem ##\dots##". If that is the case, then what does the arrow encircling the E-axis signify?
Yes, the drawing I have corresponds to the last part of the problem. I am asking about the last part of the problem as well.
One more question: The normal to the loop forms an angle of 90° with the N-axis and the N-axis forms and angle of 45° with the field. What is the angle between the field and the loop normal? What is the sine of that angle and how does it differ from the sine of 45°?
I am not 100% sure that is what the problem is saying but I suspect so. I think the angle between the field and loop normal is 90 degrees, but if that is not how you read the problem, do correct me.
 
  • #8
kuruman
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You have to understand that the plane containing the EW and NS axes is always tangent to the surface of the Earth and we call that the "horizontal plane". What we call "North" is in that plane and what we call "up" is perpendicular to the plane and away from the Earth. Therefore, "up" is everywhere perpendicular to "North" except perhaps at the poles. Here you have the loop normal in the "up" direction which makes the angle between the loop normal and "North" equal to 90°. The problem tells you that the magnetic field is "at an angle 45.0° below the horizontal." What is then the angle between the loop normal and the field?
 
  • #9
haruspex
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In my image, the loop lies in the plane of N to S and E to W.
That is still not completely clear. I assume you mean a horizontal plane, not a plane parallel to the Earth's axis.
The image is misleading then. If the loop is in the horizontal plane then the NS line should intersect it where it is furthest from the EW line. The picture makes it look as though the loop is tilted.

Into the page would be into and out of the page would be sticking out of the page.
The diagram attempts to show a 3D view from some arbitrary viewing angle. The plane of the "page" is undefined. I will assume you mean the horizontal plane.

See the black arrow below as a representation of the normal.
And I take that to mean the arrow is normal to the horizontal plane, so points vertically upward.
This normal is not at a 45 degree angle to the magnetic field
Why do you say that? The question states "Earth’s field is north, but at an angle 45.0° below the horizontal". I.e. it points down into the ground at an angle of 45° to the horizontal. If the normal is at 90° to the horizontal, and the field is at 45° to the horizontal, and both are in a vertical NS plane, what is the angle between the normal and the field?
If you are still confused about that, please post a diagram showing the view from the East as you understand it.
 
  • #10
annamal
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You have to understand that the plane containing the EW and NS axes is always tangent to the surface of the Earth and we call that the "horizontal plane". What we call "North" is in that plane and what we call "up" is perpendicular to the plane and away from the Earth. Therefore, "up" is everywhere perpendicular to "North" except perhaps at the poles. Here you have the loop normal in the "up" direction which makes the angle between the loop normal and "North" equal to 90°. The problem tells you that the magnetic field is "at an angle 45.0° below the horizontal." What is then the angle between the loop normal and the field?
Is it 90 + 45 degrees = 135 degrees?
 
  • #12
annamal
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Yes.
This is the equation for torque ##\tau = -IABsin\theta##
Plugging in the numbers ##I = -200(100)\pi(50*10^{-2})^2(6*10^{-5})sin(135) = -0.666## Amps?
 
  • #13
kuruman
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This is the equation for torque ##\tau = -IABsin\theta##
Plugging in the numbers ##I = -200(100)\pi(50*10^{-2})^2(6*10^{-5})sin(135) = -0.666## Amps?
You are looking for the magnitude and direction of the torque on the loop. Torque does not have units of Amps. Also, where does the negative sign in the equation come from?
 
  • #14
haruspex
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This is the equation for torque ##\tau = -IABsin\theta##
Plugging in the numbers ##I = -200(100)\pi(50*10^{-2})^2(6*10^{-5})sin(135) = -0.666## Amps?
... and you mean ##\tau=...##
 
  • #15
annamal
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You are looking for the magnitude and direction of the torque on the loop. Torque does not have units of Amps. Also, where does the negative sign in the equation come from?
Yes oops. yeah, not amps. I do not know how to find the direction of the torque. Could you help me?
I drew this diagram with black arrows to determine the direction of magnetic force and torque. Is that right?

When I calculate the direction of the torque = ##\vec r \times \vec F## where ##\vec r = Rcos\theta\vec i + Rsin\theta\vec j## and ##\vec F = z\vec k##; therefore torque is ##zRsin\theta\vec i - zRcos\theta\vec j##. Is that correct?
Screen Shot 2022-05-11 at 12.47.46 AM.png
 
  • #16
annamal
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Also, where does the negative sign in the equation come from?
The negative sign comes from the equation for torque in my book.
 
  • #17
haruspex
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The negative sign comes from the equation for torque in my book.
The sign only means anything in conjunction with a handedness rule. With Fleming's Left Hand Rule the sign is positive. Does the book specify its conventions?
It is easier to remember that the torque acts so as to align the loop current vector with the magnetic field, so in this case it points West.
 
  • #18
annamal
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The sign only means anything in conjunction with a handedness rule. With Fleming's Left Hand Rule the sign is positive. Does the book specify its conventions?
It is easier to remember that the torque acts so as to align the loop current vector with the magnetic field, so in this case it points West.
I am confused about that. I am pointing my fingers towards the direction of current and rolling it in the direction of B the magnetic field and my thumb points south west. So how did you get west??
 
  • #19
haruspex
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I am confused about that. I am pointing my fingers towards the direction of current and rolling it in the direction of B the magnetic field and my thumb points south west. So how did you get west??
Consider viewing the arrangement from the East.
The loop lies horizontal, the current vector points straight up and the field vector points diagonally down to the right. Both vectors lie in the plane of the "page" in this view.
To rotate the current vector to align with the field, you have to rotate it clockwise through 135°. Such a rotation is a vector into the page, i.e. West.
 
  • #20
annamal
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Consider viewing the arrangement from the East.
The loop lies horizontal, the current vector points straight up and the field vector points diagonally down to the right. Both vectors lie in the plane of the "page" in this view.
To rotate the current vector to align with the field, you have to rotate it clockwise through 135°. Such a rotation is a vector into the page, i.e. West.
Viewing from the east, if the loop is horizontal, shouldn't the current vector be pointing to the right towards north? Can you draw a picture for me?
 
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  • #21
haruspex
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Viewing from the east, if the loop is horizontal, shouldn't the current vector be pointing to the right towards north? Can you draw a picture for me?
Perhaps I should not call it the current vector. The actual current flows around the loop, not just in one direction.
I see http://labman.phys.utk.edu/phys222core/modules/m4/Current loops.html calls it the magnetic moment of the loop current. Or you can think of it as the area vector, with the current being a scalar: http://physics.bu.edu/~duffy/semester2/c13_torque.html.
 
  • #22
annamal
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Perhaps I should not call it the current vector. The actual current flows around the loop, not just in one direction.
I see http://labman.phys.utk.edu/phys222core/modules/m4/Current loops.html calls it the magnetic moment of the loop current. Or you can think of it as the area vector, with the current being a scalar: http://physics.bu.edu/~duffy/semester2/c13_torque.html.
Ok I see. If the torque vector is pointing towards the west how does the current loop rotate at all since the current loop is flat.
 
  • #23
haruspex
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Ok I see. If the torque vector is pointing towards the west how does the current loop rotate at all since the current loop is flat.
If the torque vector points to the East then it will tend to twist the loop about an EW axis, pushing the northern edge down and the southern edge up.
 
  • #24
annamal
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If the torque vector points to the East then it will tend to twist the loop about an EW axis, pushing the northern edge down and the southern edge up.
Ok, I am having trouble figuring out the direction of torque as west mathetmatically.

##dI = (Idx, Idy, 0)##
##B = (0, B_y, -B_z)##
##\vec F = \int dI \times B## = (some quantity x, y, z)
##\vec \tau = \vec r \times \vec F## = (some quantity x, y, z). If the torque were pointing to the west it would just be (-x, 0, 0)
 
  • #25
haruspex
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Ok, I am having trouble figuring out the direction of torque as west mathetmatically.

##dI = (Idx, Idy, 0)##
##B = (0, B_y, -B_z)##
##\vec F = \int dI \times B## = (some quantity x, y, z)
##\vec \tau = \vec r \times \vec F## = (some quantity x, y, z). If the torque were pointing to the west it would just be (-x, 0, 0)
So try that with ##x=r\cos(\theta)## etc.
 
  • #26
annamal
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So try that with ##x=r\cos(\theta)## etc.
What do you mean? I did try that with ##x=r\cos(\theta)##. In the end ##\vec\tau## = (some quantity x, y, z)
I forgot to mention that ##\vec r = (x, y, 0)##
 
  • #27
haruspex
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What do you mean? I did try that with ##x=r\cos(\theta)##. In the end ##\vec\tau## = (some quantity x, y, z)
I forgot to mention that ##\vec r = (x, y, 0)##
Yes, but why just "some quantity "? Fill in all the details for your outline in post #24. Everything is known.
 
  • #28
annamal
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##dI = (I(-rsin\theta d\theta), Ircos\theta d\theta, 0)##
##B = (0, B_y, -B_z)##
##\vec F = \int_{0}^{2\pi} Idl \times B## = 0
##\vec r = (rcos\theta, rsin\theta, 0)##
##\vec \tau = \vec r \times \vec F## = 0?
 
  • #29
haruspex
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##\vec F = \int_{0}^{2\pi} Idl \times B## = 0
That integral makes no sense. You show dl as the integration variable but the range seems to be for an angle. Write it all in terms of ##r, \theta## instead of l and perform that cross product.
 
  • #30
haruspex
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##dI = (I(-rsin\theta d\theta), Ircos\theta d\theta, 0)##
Actually that line doesn't make sense either. The LHS is a current element but the RHS is a current times a length element.
 
  • #31
annamal
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Actually that line doesn't make sense either. The LHS is a current element but the RHS is a current times a length element.
I meant
##Idl = (I(-rsin\theta d\theta), Ircos\theta d\theta, 0)##

unsure why dl has to be in terms of r and ##\theta##
if so, does ##\vec dl = rd\theta (-\vec i + \vec j)##
 
  • #32
annamal
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That integral makes no sense. You show dl as the integration variable but the range seems to be for an angle. Write it all in terms of ##r, \theta## instead of l and perform that cross product.
dI = (dx, dy, 0) = ##(-rsin\theta d\theta), rcos\theta d\theta, 0)##
where x = ##rcos\theta## and y = ##rsin\theta##
 
  • #33
haruspex
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why dl has to be in terms of r and θ
So that you can do the integral.
if so, does ##\vec dl = rd\theta (-\vec i + \vec j)##
No, it is what you wrote,
##(-rsin\theta d\theta, rcos\theta d\theta, 0)##
Now substitute that for ##\vec{dl}## in ##\vec F = \int_{\theta=0}^{2\pi} I\vec{dl} \times \vec B##, and replace ##\vec B## there using ##\vec B = (0, B_y, -B_z)##. Then perform the cross product, then perform the integration.
 
  • #34
annamal
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So that you can do the integral.

No, it is what you wrote,

Now substitute that for ##\vec{dl}## in ##\vec F = \int_{\theta=0}^{2\pi} I\vec{dl} \times \vec B##, and replace ##\vec B## there using ##\vec B = (0, B_y, -B_z)##. Then perform the cross product, then perform the integration.
Ok, so ##(-Irsin\theta d\theta, Ircos\theta d\theta, 0) \times (0, B_y, -B_z) = \vec dF = (-IB_z rcos\theta d\theta, -IB_z rsin\theta d\theta, -IB_y rsin\theta d\theta)##
##\vec F = (-IB_z rsin\theta, IB_z rcos\theta, IB_y rcos\theta)## from 0 to ##2\pi## = 0?
 
  • #35
haruspex
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Ok, so ##(-Irsin\theta d\theta, Ircos\theta d\theta, 0) \times (0, B_y, -B_z) = \vec dF = (-IB_z rcos\theta d\theta, -IB_z rsin\theta d\theta, -IB_y rsin\theta d\theta)##
##\vec F = (-IB_z rsin\theta, IB_z rcos\theta, IB_y rcos\theta)## from 0 to ##2\pi## = 0?
Ok, so you found the net force is zero. But we don't care about the net force, we care about the net torque. So find the torque on an element and then integrate.
 

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