- #1
zenterix
- 708
- 84
- Homework Statement
- This question is about considering a magnetic dipole moment located at the origin of a coordinate system under effect of a magnetic field ##\vec{B}##. For different definitions for ##\vec{B}##, we need to find the net torque and the resultant force.
- Relevant Equations
- This is a problem from MIT OCW's online course 8.02. There is an automated solution system, but it only tells me what is wrong, not why. Thus, I spent an enormous amount of time today on this one problem without making too much progress.
Consider the following scenario.
We have a closed current loop (say a rectangular loop to make it easier) with magnetic dipole moment ##\vec{\mu}## shown below.
We have a magnetic field ##\vec{B}## which will be defined in different ways below.
I think it is easier to visualize as follows
Where the blue line represents a side view of the current loop.
Let
$$\vec{\mu}=\mu_C(\hat{i}+\hat{j})$$
where ##\mu_C## is a positive constant.
Consider the following cases for the definition of magnetic field ##\vec{B}##
a) ##\vec{B}=B_0\hat{j}##
b) ##\vec{B}=-B_0(\hat{i}+\hat{j}##
c) ##\vec{B}=B_0\frac{x}{d}\hat{i}##
d) ##\vec{B}=-B_0\frac{x}{d}\hat{j}##
e) ##\vec{B}=B_0\frac{x^2}{d^2}\hat{j}##
f) ##\vec{B}=B_0\left (\frac{x_0}{d}-\frac{x^2}{d^2}\right )\hat{j}##
where ##x_0## and ##d## are positive constants in units of meters and ##x## is the ##x## component of a point on the ##xy##-plane.
The dipole is fixed so it cannot rotate or move.
We want to know the net torque and the resultant force.
Hence, essentially, we repeat the same analysis once for each of the items.
I thought this would be as simple as computing ##\vec{\mu}\times \vec{B}## in each case to obtain torque.
However, I think this only works in cases a) and b) since in the other cases the magnetic field depends on ##x##.
Now, the calculations below are probably way overkill relative to what is required in this problem. However, I still think they should produce results that can answer this problem, albeit in a circuitous way. Yet, what I do below does not seem to produce the correct answers (at least according to the grading system).
One question I have is how to solve the problem in a simpler way using perhaps the given ##\vec{\mu}##?
At some point I decided to do the calculations assuming that the current loop is rectangular.
The blue line in the picture above would then be one of the sides, and there would be two sides perpendicular to the screen (one with current going into the screen and the other with current coming out of the screen).
Here is another view
As far as I can tell, for this problem (in which ##\vec{B}## is always in the ##xy##-plane) the only sides that contribute to net torque are (a) and (c) above. The forces on (b) and (d) cancel out in terms of torque and in terms of just their sum. Is this actually true? I did not try to do actual calculations, I simply contorted my right hand to figure out the right hand rule cross product for each piece of (b) and (d) in each case for ##\vec{B}##.
For each case, I carried out the calculation
$$d\vec{F}_a(x,-x,z)=Id\vec{s}_a\times \vec{B}(x,-x,z)=Idz(-\hat{k}) \times\vec{B}(x,-x,z)$$
$$d\vec{F}_c(-x,x,z)=Id\vec{s}_c\times \vec{B}(-x,x,z)=Idz \hat{k} \times\vec{B}(-x,x,z)$$
$$\vec{F}_a= \int_a d\vec{F}_a=\int_{-l}^l \left (Idz(-\hat{k})\right )\times\vec{B}(x,-x,z)$$
$$\vec{F}_c=\int_c d\vec{F}_c=\int_{-l}^l (Idz\hat{k})\times\vec{B}(-x,x,z)$$
and then
$$\vec{\tau}=\int_a \vec{r}_a \times \vec{F}_a+\int_c \vec{r}_c\times\vec{F}_c$$
$$=\int_{-l}^l (x\hat{i}-x\hat{j}+z\hat{k})\times \vec{F}_a=\int_{-l}^l (-x\hat{i}+x\hat{j}+z\hat{k})\times\vec{F}_c$$
In other words, I compute the torque due to sides (a) and (c) be integrating the infinitesimal torques due to each piece of such sides.
Following are the results for each case.
Note that in each case, I am showing an array. The entries from top to bottom are ##\vec{F}_a, \vec{F}_c, \vec{F}_a+\vec{F}_c##, and ##\vec{\tau}##.
Note that ##x>0## and that ##x## is a constant.
The questions are
1) In which case (or cases) will the dipole experience a torque that tends to rotate it counterclockwise as viewed from ##+z##?
2) In which case (or cases) will the dipole experience a torque that tends to rotate it clockwise as viewed from ##+z##?
3) In which case (or cases) will the dipole experience no torque?
4) In which case (or cases) will the dipole experience a force that has a component in the ##+x## direction?
5) In which case (or cases) will the dipole experience a force that has a component in the ##-x## direction?
6) In which case (or cases) will the dipole experience a force that has a component in the ##+y## direction?
7) In which case (or cases) will the dipole experience a force that has a component in the ##-y## direction?
8) In which case (or cases) will the dipole experience no force?
My answer to these questions for the specific case of a rectangular loop are
1) (a) and (e). Also, (f) if ##x^2<dx_0##.
2) Only (f) when ##x^2>dx_0##.
3) (b), (c), (d)
4) None
5) (d)
6) None
7) (c)
8) (a), (b), (e), (f)
The solution system says 1), 3), 4), and 7) are incorrect.
We have a closed current loop (say a rectangular loop to make it easier) with magnetic dipole moment ##\vec{\mu}## shown below.
We have a magnetic field ##\vec{B}## which will be defined in different ways below.
I think it is easier to visualize as follows
Where the blue line represents a side view of the current loop.
Let
$$\vec{\mu}=\mu_C(\hat{i}+\hat{j})$$
where ##\mu_C## is a positive constant.
Consider the following cases for the definition of magnetic field ##\vec{B}##
a) ##\vec{B}=B_0\hat{j}##
b) ##\vec{B}=-B_0(\hat{i}+\hat{j}##
c) ##\vec{B}=B_0\frac{x}{d}\hat{i}##
d) ##\vec{B}=-B_0\frac{x}{d}\hat{j}##
e) ##\vec{B}=B_0\frac{x^2}{d^2}\hat{j}##
f) ##\vec{B}=B_0\left (\frac{x_0}{d}-\frac{x^2}{d^2}\right )\hat{j}##
where ##x_0## and ##d## are positive constants in units of meters and ##x## is the ##x## component of a point on the ##xy##-plane.
The dipole is fixed so it cannot rotate or move.
We want to know the net torque and the resultant force.
Hence, essentially, we repeat the same analysis once for each of the items.
I thought this would be as simple as computing ##\vec{\mu}\times \vec{B}## in each case to obtain torque.
However, I think this only works in cases a) and b) since in the other cases the magnetic field depends on ##x##.
Now, the calculations below are probably way overkill relative to what is required in this problem. However, I still think they should produce results that can answer this problem, albeit in a circuitous way. Yet, what I do below does not seem to produce the correct answers (at least according to the grading system).
One question I have is how to solve the problem in a simpler way using perhaps the given ##\vec{\mu}##?
At some point I decided to do the calculations assuming that the current loop is rectangular.
The blue line in the picture above would then be one of the sides, and there would be two sides perpendicular to the screen (one with current going into the screen and the other with current coming out of the screen).
Here is another view
As far as I can tell, for this problem (in which ##\vec{B}## is always in the ##xy##-plane) the only sides that contribute to net torque are (a) and (c) above. The forces on (b) and (d) cancel out in terms of torque and in terms of just their sum. Is this actually true? I did not try to do actual calculations, I simply contorted my right hand to figure out the right hand rule cross product for each piece of (b) and (d) in each case for ##\vec{B}##.
For each case, I carried out the calculation
$$d\vec{F}_a(x,-x,z)=Id\vec{s}_a\times \vec{B}(x,-x,z)=Idz(-\hat{k}) \times\vec{B}(x,-x,z)$$
$$d\vec{F}_c(-x,x,z)=Id\vec{s}_c\times \vec{B}(-x,x,z)=Idz \hat{k} \times\vec{B}(-x,x,z)$$
$$\vec{F}_a= \int_a d\vec{F}_a=\int_{-l}^l \left (Idz(-\hat{k})\right )\times\vec{B}(x,-x,z)$$
$$\vec{F}_c=\int_c d\vec{F}_c=\int_{-l}^l (Idz\hat{k})\times\vec{B}(-x,x,z)$$
and then
$$\vec{\tau}=\int_a \vec{r}_a \times \vec{F}_a+\int_c \vec{r}_c\times\vec{F}_c$$
$$=\int_{-l}^l (x\hat{i}-x\hat{j}+z\hat{k})\times \vec{F}_a=\int_{-l}^l (-x\hat{i}+x\hat{j}+z\hat{k})\times\vec{F}_c$$
In other words, I compute the torque due to sides (a) and (c) be integrating the infinitesimal torques due to each piece of such sides.
Following are the results for each case.
Note that in each case, I am showing an array. The entries from top to bottom are ##\vec{F}_a, \vec{F}_c, \vec{F}_a+\vec{F}_c##, and ##\vec{\tau}##.
Note that ##x>0## and that ##x## is a constant.
The questions are
1) In which case (or cases) will the dipole experience a torque that tends to rotate it counterclockwise as viewed from ##+z##?
2) In which case (or cases) will the dipole experience a torque that tends to rotate it clockwise as viewed from ##+z##?
3) In which case (or cases) will the dipole experience no torque?
4) In which case (or cases) will the dipole experience a force that has a component in the ##+x## direction?
5) In which case (or cases) will the dipole experience a force that has a component in the ##-x## direction?
6) In which case (or cases) will the dipole experience a force that has a component in the ##+y## direction?
7) In which case (or cases) will the dipole experience a force that has a component in the ##-y## direction?
8) In which case (or cases) will the dipole experience no force?
My answer to these questions for the specific case of a rectangular loop are
1) (a) and (e). Also, (f) if ##x^2<dx_0##.
2) Only (f) when ##x^2>dx_0##.
3) (b), (c), (d)
4) None
5) (d)
6) None
7) (c)
8) (a), (b), (e), (f)
The solution system says 1), 3), 4), and 7) are incorrect.
Last edited: