Lorentz force is F=q(E+v cross B).................................................. .................(1) We seem to be interested only in B[=A(q/r^2)(v' cross ].What about E?Is it an electrostatic field?I suppose not.If not,then should be time dependent and del cross E=-(d/dt)B Taking line integral of (1),W=integral(a to b)F.dr =integral(a to b)E.dr + 0 Does this mean F(mag) is conservative? if a and b are the same,Will W=0? in that case will Lorentz force be conservative? however, i saw in Griffiths's Quantum Mechanics that magnetic forces cannot be expressed like (-dV/dx) like other conservative forces. what is the physics?
A conservative force is one for which the curl is zero. i.e. force F is conservative if and only if: \(\nabla \times F = 0.\) This is true for electrostatic forces. However, if at a point in space the magnetic field is changing in time, then the electric force will, in general, be non-conservative. The changing magnetic field induces a curl in the electric field. The Lorentz force is \(F = qE + qv \times B\) Therefore: \(\nabla \times F = q(\nabla \times E - B(\nabla.v) + (B.\nabla)v - (v.\nabla)B)\) As you can see, in general this will not be equal to zero.
from your post it's also clear F(mag)=q(v cross B) is also non-conservative,in general.then why dW=intgral F(mag) . dr always vanishes?What is the physical reason in behind?
Hi neelakash, I can't answer your question, but I can show you a trick. The [thread=61223] Displaying equations using Tex[/thread] thread shows how you can insert pretty equations into your posts. For example, you can do F(mag)=q(v cross B) like this: \(F_{mag} = q(v \times B)\)To get this: \(F_{mag} = q(v \times B)\) And dW=intgral F(mag) like this: \(dW = \int F_{mag}.dr\)To get this: \(dW = \int F_{mag}.dr\) Also, if you click the quote button on any post that uses tex codes (like James's posts in this thread), you can see the code used and borrow it for your own posts.
Yes, if you take dr along the path of the particle. F(mag) is always perpendicular to dr, so the integral must be zero.
But the definition of a conservative force is that the work done along any path between two points is independent of the path chosen. Hmm... I think I need to think about this some more.
Yes!Please think over it. And responding to Pete's post, where to write in that way?And After I write the entire thing, what should I do?such as pressing ENTER, or,clicking somewhere...
Hi neelakash, Just type and post it the same as you usually do. Try typing this as part of a post, and see what happens: \(W =int_a^bF.dr\)
Using a bit more vector calculus, and realising that \(v = \frac{dr}{dt}\) we find that \(\nabla \times F = -q(\frac{dB}{dt} + (v.\nabla)B)\) where F is the Lorentz force, v is the velocity and B is the magnetic field. For the force to be conservative, we need curl F = 0. Obviously, curl F won't be zero in general. It seems to me that it will only be zero in a uniform magnetic field (which might itself be zero) that is unchanging in time. The problem, I think, is that the magnetic force is velocity dependent, rather than merely being position dependent. Even this isn't a complete answer, since I haven't yet sorted out in my own mind how the magnetic force can do work when it is always perpendicular to the instantaneous velocity. Perhaps the problem is that in an inhomogenous magnetic field, if you take a charged particle from point A to point B along two different paths, the magnetic force will, in general, do different amounts of work on the particle. This is why the second term appears in the expression for the curl, above. The first term appears if the magnetic field also varies in time.
Well,I will think more while continuing the discussion.But, in response to your sayings in the last part, magnetic forces never work on moving charges.
