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location Kolkata,India[91-33-25514464]
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visits member for 2 years, 9 months
seen Mar 27 '13 at 12:06

Author/Teacher from India. Interested in General Relativity and other areas of physics


Aug
14
comment How does one prove:$\nabla(\vec{\mu_m}\cdot\vec{B})\cdot\vec{dr}=0$?
Magnetic force cannot perform work but electric field can do work.Such a field may result from the spatial variation of B from the curl B equation[comment to the answer may be considered]
Aug
14
comment How does one prove:$\nabla(\vec{\mu_m}\cdot\vec{B})\cdot\vec{dr}=0$?
Spatial variation of $\vec{B}$ can produce an electric field:$\nabla \times \vec{B}=\mu_0 \vec{j}+\mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}$.If $\vec{E}\cdot\vec{dr}=\nabla(\vec{\mu_m}\cdot\vec{B})\cdot\vec{dr}$ to produce non-zero work then,$\vec{E}=\nabla(\vec{\mu_m}\cdot\vec{B})$. How do you prove that by direct calculation?[here you have a view of the problem through the differential equations which are supposed to represent the physical conditions accurately.]
Aug
13
comment How does one prove:$\nabla(\vec{\mu_m}\cdot\vec{B})\cdot\vec{dr}=0$?
Let the magnetic field be in the x direction.If the observer moves along the y direction with a uniform speed v,we have $\vec{E}=\gamma(\vec{v}\times\vec{B})$. The electric field accelerates the charged changing its speed and doing work. Initially there was a acceleration normal to the direction of motion. But now we have acceleration due to change of speed also . If the net acceleration changes the radiating power should be different from the two frames.But energy is not an invariant wrt frame transformation though it is a conserved quantity. It might have different values in differentframes
Aug
13
comment How does one prove:$\nabla(\vec{\mu_m}\cdot\vec{B})\cdot\vec{dr}=0$?
You can think of redistribution of charges and currents elsewhere(gadgets producing the field) to produce an electric field due to variations of magnetic field at some fixed point. But a pure magnetic field which does not vary with time cannot perform any work.What would be your mechanism for accounting for such variations of $\vec{B}$ in evaluating:$ \nabla (\vec{\mu_m}.\vec{B}).\vec{dr}$. What would be the mechanism of including electric field in the said evaluation of elementary work?
Aug
13
asked How does one prove:$\nabla(\vec{\mu_m}\cdot\vec{B})\cdot\vec{dr}=0$?
Aug
12
comment The Stern Gerlach Experiment Revisited
In so far as the last equation in the answer is concerned, the value of x is the same for all particles passing though the region of interaction of the magnetic field. The value of y is confined to the dimemsions of the region of interaction. Its becomes negligible due to the large vale of $\frac{C_1}{k}$. We have in effect:$\frac{C_1}{k}z+\frac{c_2}{k}=(V\pm v_m \frac{C_1}{k})t$
Aug
12
answered The Stern Gerlach Experiment Revisited
Aug
11
revised The Stern Gerlach Experiment Revisited
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Aug
11
revised The Stern Gerlach Experiment Revisited
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Aug
11
revised The Stern Gerlach Experiment Revisited
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Aug
11
revised The Stern Gerlach Experiment Revisited
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Aug
11
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Aug
11
awarded  Critic
Aug
11
comment The Stern Gerlach Experiment Revisited
We may take $\langle v_x \rangle=\langle v_y \rangle$. They correspond to internal motion.The averages masy be taken over time or space and they should be equal in consideration of the Ergodicity principle
Aug
11
revised The Stern Gerlach Experiment Revisited
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Aug
11
comment The Stern Gerlach Experiment Revisited
Do you find anything counter-intuitive with the above considerations(ie, the answer provided)?Classical prejudice seems to be working out excellently ,at least with the problem at hand---the Stern-Gerlach Experiment.
Aug
11
revised The Stern Gerlach Experiment Revisited
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Aug
11
answered The Stern Gerlach Experiment Revisited
Aug
10
comment The Stern Gerlach Experiment Revisited
$\vec{F}=\nabla(\vec{\mu}.\vec{B})\approx\vec{e}_{z}\mu_{z}\frac{\partial B_{z}}{\partial z}=\vec{e}_{z}F_{z}$ Force corresponding to the above formula acts along the direction of $\vec{B}=\vec{e_z}B_z$, if $B_y=B_z=0$ The above force is not identical with the force given by $F=q(\vec{v}\times \vec{B})$. Magnetic force cannot act along the magnetic line of force
Aug
10
comment The Stern Gerlach Experiment Revisited
Classical theory fails to predict the "excluded middle". This goes in conjunction with the "classic" exclusion or unaccountability of how spin generates retardation in the x-y plane if $B_x=B_y=0$. Work done by magnetic field for any infinitesimal path is zero. This idea is there in the classical theories and needs to be respected in any type of classical reasoning.