Problem related to the vector potential of B

In summary, the particle moves in a magnetic field and its direction is determined by the change in its position and the magnetic field.
  • #1
nuclear_dog
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0

Homework Statement


A particle of mass m and charge q moves in a region with a magnetic field B(r) which is time independent. What is [itex]\frac{dA}{dt}[/itex] as seen by the particle.

Homework Equations


The Attempt at a Solution


Since the B field is not varying with time, I know that the change in A will be due to the change in position of particle with time. It should be (∂A/∂x)vx + (∂A/∂y)vy + (∂A/∂z)vz . So can it be (v.∇)A. But I cannot seem to be able to go furthur.
 
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  • #2
A is a scalar, so (v.∇)A n= v.(∇A), right? And ∇A =?
 
  • #3
I am sorry I forgot to add that in the question it was given that since ∇.B = 0 , therefore we can write B = ∇XA , where A is a vector. So I am not sure that we can use the transformation of ∇ in the expression.
 
  • #4
OK, I see. What makes you think (v.∇)A is an inadequate answer?
 
  • #5
haruspex , I got hold of the original question paper today and have attached a screenshot of it. Please take a look at it to clearly understand the question , since I might not have explained it properly.

Since (dp/dt) = F = q(vXB) , hence if we can show that ∇X{d/dt( p + qA)} = 0 , it would mean that (d/dt(p + qA)) is a gradient of some function. But I haven't been able to prove that ∇X{d/dt( p + qA)} = 0 , by using dA/dt = (v.∇)A .
 

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  • #6
##\frac{d\vec{p}}{dt} =q\vec{v}\times\vec{B}##
##\frac{d}{dt}\left(\vec{p}+q \vec{A}\right) = q \vec{v}\times \vec{B}+q \left( \vec{v}.\vec{\nabla} \right) \vec{A} = q \vec{v}\times \left( \vec{\nabla}\times\vec{A}\right)+q \left( \vec{v}.\vec{\nabla} \right) \vec{A} ##
Here I venture into areas I know even less about, based what I read at http://en.wikipedia.org/wiki/Curl_%28mathematics%29:
## \vec{v}\times \left(\vec{\nabla}\times\vec{A}\right) = \vec{\nabla_A}\left(\vec{v}.\vec{A}\right) - \left(\vec{v}.\vec{\nabla}\right)\vec{A}##
So
##\frac{d}{dt}\left(\vec{p}+q \vec{A}\right) = q \vec{\nabla_A}\left(\vec{v}.\vec{A}\right)##
Does that help?
 
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  • #7
Thanks, that actually solves the problem. Although I will have to read up on the identities of vector calculus to feel satisfied on this question.
 

FAQ: Problem related to the vector potential of B

1. What is the vector potential of B?

The vector potential of B is a mathematical quantity that is used to describe the magnetic field in a given space. It is a vector field that is related to the magnetic field by a mathematical operation called the curl.

2. How is the vector potential of B related to the magnetic field?

The vector potential of B is related to the magnetic field by the equation B = ∇ x A, where B is the magnetic field, A is the vector potential, and ∇ x is the curl operator. This equation is known as the Biot-Savart law.

3. What are the units of the vector potential of B?

The units of the vector potential of B depend on the system of units being used. In SI units, the vector potential is measured in tesla-meters (Tm). In cgs units, it is measured in gauss-centimeters (Gcm).

4. What is the physical significance of the vector potential of B?

The vector potential of B has no direct physical significance, but it is a useful mathematical tool for describing and calculating the behavior of magnetic fields. In electromagnetism, it is used to calculate the magnetic field from a given distribution of electric currents.

5. How is the vector potential of B used in practical applications?

The vector potential of B is used in many practical applications, such as in the design of electromagnets and magnetic sensors. It is also used in the field of electromagnetism to calculate the magnetic field of a moving charged particle, as well as in the study of electromagnetic waves and their interaction with matter.

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