# Homework Help: Force on a dipole moving through a magnetic field

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1. Mar 1, 2016

### pondzo

1. The problem statement, all variables and given/known data

A dipole of moment $\vec{p}$ where p is fixed, moves with velocity $\vec{v}$ though a magnetic field $\vec{B}$. Show that the force on the dipole is $\vec{v}\times(\vec{p}\cdot\vec{\nabla})\vec{B}+\dot{\vec{p}}\times\vec{B}$.

2. Relevant equations

$\vec{F}=e(\vec{E}+\vec{v}\times\vec{B})$

3. The attempt at a solution
This is how I am picturing the situation:

$\vec{F}=\sum_i\vec{F_i}$
$~~~=e\vec{v}\times\vec{B_1}-e\vec{v}\times\vec{B_2}$
$~~~=e\vec{v}\times\vec{B}(\vec{a}+\frac{1}{2}\vec{d})-e\vec{v}\times\vec{B}(\vec{a}-\frac{1}{2}\vec{d})$, Now I taylor expand $\vec{B}$ and disregard high order terms:
$~~~\approx e\vec{v}\times[\vec{B}(\vec{a})+\frac{1}{2}\vec{d}\cdot\vec{\nabla}\vec{B}|_{\vec{a}}]-e\vec{v}\times[\vec{B}(\vec{a})-\frac{1}{2}\vec{d}\cdot\vec{\nabla}\vec{B}|_{\vec{a}}]$
$~~~=e\vec{v}\times\vec{B}(\vec{a})+\frac{1}{2}e\vec{v}\times(\vec{d}\cdot\vec{\nabla}\vec{B}|_{\vec{a}})-e\vec{v}\times\vec{B}(\vec{a})+\frac{1}{2}e\vec{v}\times(\vec{d}\cdot\vec{\nabla}\vec{B}|_{\vec{a}})$
$~~~=e\vec{v}\times(\vec{d}\cdot\vec{\nabla}\vec{B}|_{\vec{a}})$
$~~~=\vec{v}\times(\vec{p}\cdot\vec{\nabla}\vec{B}|_{\vec{a}})$
Using the identity $\vec{\nabla}\times(\phi\vec{B})=(\vec{\nabla}\phi)\times\vec{B}+\phi\vec{\nabla}\times\vec{B}$ and replacing $\vec{\nabla}$ with $\vec{v}$, also replacing $\phi$ with $(\vec{p}\cdot\vec{\nabla})$ we get:

$~~~=\vec{v}(\vec{p}\cdot\vec{\nabla})\times\vec{B}+(\vec{p}\cdot\vec{\nabla})\vec{v}\times\vec{B}$
$~~~=\vec{v}\times (\vec{p}\cdot\vec{\nabla})\vec{B}+(\vec{p}\cdot\vec{\nabla})\vec{v}\times\vec{B}$

I think the above derivation is correct so far, so all that remains is to show that $(\vec{p}\cdot\vec{\nabla})\vec{v}\equiv \dot{\vec{p}}$ But I am stuck with this, any help will be appreciated.

2. Mar 1, 2016

### haruspex

The first of those does not look like a valid substitution to me, and I'm not sure about the second. The formula depends on the differential nature of $\nabla$.
I would certainly not expect to see a $\dot v$ term arise, and it would not be related to $\dot p$. You are given that the dipole is constant in magnitude. What does variation in the vector over time therefore represent physically?

3. Mar 1, 2016

### pondzo

Yes I was quite iffy about making this substitution, do you know of any identity I can apply in order to expand this expression?

I think it is supposed to represent any rotation the dipole might be experiencing in the field. Although i'm not sure what bearing p being fixed has on the outcome of this question.

So my last two lines of working are incorrect which is due to the dodgy substitution I made?

4. Mar 1, 2016

### haruspex

Yes, I agree with all of that. Not sure of the correct path though... will try to spend some more time on it later.

5. Mar 1, 2016

### pondzo

Thank you, I will post if I get any farther.

6. Mar 1, 2016

### haruspex

Ok.
I tried thinking of the dipole as two charges, so a rotation of the dipole looks like two parallel currents in the same direction. When the dipole is aligned with the magnetic field, I can see that this produces a net force; but when the dipole, the field, and the rotation vector of the dipole are mutually orthogonal the two currents are aligned with the field, so I don't see why there is a force.

7. Mar 1, 2016

### TSny

I think you should let each charge have its own velocity: $\vec{v_1}$ and $\vec{v_2}$.
Shouldn't $\vec{B}(\vec{a}+\frac{1}{2}\vec{d}) = \vec{B}(\vec{a}) + \frac{1}{2} ( \vec{d}\cdot\vec{\nabla} ) \vec{B}|_{\vec{a}}$? (with parentheses)

8. Mar 1, 2016

### pondzo

But won't the current be in the direction in which the charges are moving,$\vec{v}$ ? which would be perpendicular to the magnetic field, so it would feel a force?

Ok and then would it be the case that $\vec{v_1}=-\vec{v_2}$ ?

Yeah that should be the case, but does this change my answer (prior to the last two lines)?

9. Mar 1, 2016

### haruspex

Say the field is along the z axis, the dipole is along the x axis, and the rotation of the dipole along the y axis. Ignoring the linear movement of the dipole, the charges are moving parallel to the z axis, i.e. the currents are parallel to the field.
Anyway, since TSny has joined the thread you are in much more capable hands, so I shall bow out.

10. Mar 1, 2016

### TSny

No. Their sum is related to $\dot{\vec{a}}$.

I don't think you need to invoke the vector identity that you used in the first post. Yes, you should get your final result plus a term that comes from the particles having different velocities.

11. Mar 1, 2016

### TSny

Please don't. Excuse my jumping in.
I just happened to find that the result follows once you assign different velocities to the two particles.

12. Mar 1, 2016

### TSny

I'm probably doing something wrong since I get the stated result without assuming that $|\vec{p}|$ is fixed. I'll need to check over my work.

13. Mar 1, 2016

### pondzo

Ok I think I have got it. Please excuse my laziness but I would prefer not typesetting it and just posting a picture.

14. Mar 1, 2016

### TSny

That looks good. Note that the condition "p is fixed" is not used.