# Non-conservative or not?

1. Mar 31, 2008

### i_island0

magnetic field forms closed loop always. So, it should be called a non-conservative force.

But if a charged particle enters a magnetic field, it would move in a straight line or circular path or an helical path. In all the three cases, the magnetic field wont do any work on the charged particle. So, is it worth talking about the non-conservative in nature of magnetic field.

2. Mar 31, 2008

### rohanprabhu

A non-conservative force is a force which does zero work in a closed loop in simple terms means that, if an object under a force is moved a path AB and the work done be 'W', then if the object is moved along the path BA keeping all under conditions same, then the work done by the force will be '-W'. Mathematically, you can say that:

$$\int_A^{B} Fdr + \int_B^{A} Fdr = 0$$

for a conservative force 'F'. The 'closed loop' that we talk about isn't the loop made by the field.. but rather the path moved by an object under that force.

The direction of a field, in a sense is arbitary. The direction of Magnetic field is defined as 'the direction along the component of velocity on which the force applied does not depend upon'.

The work turns out to be 0 since, the Force is always perpendicular to the displacement. Take the example when the charge makes a circle. The Force direction is radial whereas the displacement is tangential and hence the cross product F.dr = 0.

However, a magnetic field does work on a magnetic dipole. A magnetic dipole in a magnetic field has an associated potential energy. Also, if you move a magnetic dipole along paths AB and BA in a magnetic field, the work done on them is 0. Magnetic force is a conservative force.

3. Mar 31, 2008

### i_island0

hey, can you put some more light on this. I read somewhere that magnetic field is non-conservative field.
Can it be proven mathematically that magnetic field [uniform or non-uniform] is a conservative field always.

4. Mar 31, 2008

### nicksauce

Easily... Work is the path integral of force dotted with velocity. Magnetic force is given by the cross product of velocity and magnetic field. Therefore Magnetic force is perpendicular to velocity. Therefore Magnetic force dotted with velocity is 0. Therefore work is 0. (Here I am assuming conservative <-> does no work in a closed path (or in this case any path))

5. Apr 1, 2008

### i_island0

ok fine.. but the general definition of a non-conservative field lines is that they form closed loop.
So, what would shall we conclude, in general, about the nature of magnetic field. Conservative or non-conservative.

6. Apr 1, 2008

### BryanP

Again, a field is conservative if the work done around an arbitrary curve in the force field is zero throughout the closed path C.

In this case the magnetic force is perpendicular to the velocity throughout the integral C and this results in a dot product yielding zero.

The point is if the closed loop integral's work is zero, then the field is conservative.

You can read a summary of conservative forces here: http://en.wikipedia.org/wiki/Conservative_force

I think this excerpt may be of value to you:

Last edited: Apr 1, 2008
7. Apr 1, 2008

### Hootenanny

Staff Emeritus
The magnetic field is a conservative vector field since one may write it as the gradient of a scalar field, the potential field,

$$\underline{B} = \nabla A\left(\underline{p}\right)$$

Therefore by definition the magnetic field is conservative.

8. Apr 1, 2008

### i_island0

thx... i got few more questions in this regard. IF you can tell me if I am conceptually correct.
Electric field produced by stationary charges - conservative in nature

Induced electric field - non-conservative nature

Uniform Magnetic field (forming closed loops) - conservative in nature

Non- Uniform Magnetic field (forming closed loops) - conservative in nature

magnetic field Changing with time- ??? Shall i call it conservative or since its inducing and electric field.. so rather call it non-conservative.

9. Apr 1, 2008

### Hootenanny

Staff Emeritus
Correct
A time-varying magnetic field is still conservative (since it still satisfies the condition I detailed previously). However, as you correctly say, the electric field induced by a time-varying magnetic field is non-conservative.

10. Apr 1, 2008

### i_island0

thats great.. thx.. i got it this time

11. Apr 1, 2008

### nicksauce

Suppose $$\vec{B} = \nabla{A}$$
Then
$$\nabla\times\vec{B} = \nabla\times\nabla{A}$$
Now the curl of a gradient is always 0, so
$$\nabla\times\vec{B} = 0 = \mu_0\vec{J}$$

So this is only true in a current free region. Thus I would say the magnetic field is only conservative in the sense that it does no work in a closed path, but not in the sense that there exists a scalar potential A such that B = grad A. In standard vector calculus you learn that these statements are equivalent, but that assumes the vector functions depend only on position, and not on path, as does the magnetic force.

Last edited: Apr 1, 2008
12. Apr 1, 2008

### Hootenanny

Staff Emeritus
You are of course correct nick, the scalar potential is not defined where there exists a non-zero current density. One can allow sources in the magnetic potential, provided that the sources lie on the discontinuities of the potential.

I should also add however, that we can always write the magnetic field as the curl of the magnetic vector potential, that is we can always write,

$$\underline{B} = \text{curl}\left(\underline{A}\right)$$

Last edited: Apr 1, 2008
13. Apr 1, 2008

### i_island0

So, if I make a unit magnetic monopole move in a closed path around a current carrying wire, there is some non-zero work done. Then in that case, can we say that magnetic field is non-conservative in nature.??