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lorentz force

 
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Aug29-05, 01:07 AM   #1
 

lorentz force


hi,
this is a challenging question
there are two charges q1 and q2 .At some instant of time first one is moving towards the second one with a uniform speed v1 and q2 is moving perpendicular to the line joining q1 and q2 with a speed v2. calculate forces between them ie F12 and F21. Are they equal and opposite .if not what happens to newtons third law
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Aug29-05, 01:43 AM   #2
 
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lorentz force implies a magnetic field... is there no mention of a magnetic field in the question?
Aug29-05, 03:50 PM   #3
 
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Sure, Quasar - there's a non-zero B-field around Q2 at the location of Q1,
and there's a changing E-field at the location of Q2 caused by moving Q1.

Great Question! Thanks, Shahidkilji
Aug29-05, 05:12 PM   #4
 
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lorentz force


around Q2? What does it look like exactly? What equation gives the form of this field?
Aug29-05, 08:15 PM   #5
 
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The B-field surrounds a moving charge, transverse to its velocity.
Biot-Savart Law has B = (mu_0/4pi) (q v_vec x r_hat)/r^2 .
The cross-product leads to |B| = (mu_0/4pi)(qv/r^2)sin(theta)
(right-hand-thumb along qv_vec, fingers wrap along B-field),
with theta = the polar angle from the forward direction (v_vec).
Roughly, B looks like loops encircling the charge as it moves.

(To purists who may object that Biot-Savart is incorrect:
(this is the *College HW* forum! Yes, Biot-Savart is written
(this way so that Ampere's Law can be derived from it, and
(of course Ampere's Law must be augmented with a dB/dt term.
(That's why I thought this was a great question!

So, using Biot-Savart only, B (at location 2 by q1) = 0
while B (at location 1 caused by q2) is nonzero.
Aug30-05, 12:06 AM   #6
 
so u have concluded that mag field due to q1 on q2 is zero but vice versa is not true . therefore lorentz force due to q1 on q2 will be zero but due to q2 on q1 will not be zero . Is it not violation of newtons third law.
Aug30-05, 02:43 AM   #7
 
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magnetic field at q1 due to q2 is not constant. did you accounted the electric field at q1 due to changing magnetic field of q2 at q1?
Aug30-05, 10:23 PM   #8
 
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Quote by lightgrav
The B-field surrounds a moving charge, transverse to its velocity.
Biot-Savart Law has B = (mu_0/4pi) (q v_vec x r_hat)/r^2 .
The cross-product leads to |B| = (mu_0/4pi)(qv/r^2)sin(theta)
(right-hand-thumb along qv_vec, fingers wrap along B-field),
with theta = the polar angle from the forward direction (v_vec).
Roughly, B looks like loops encircling the charge as it moves.
I never saw Biot-Savard law written that way. On the other hand, Griffiths pp.215 says that the Biot-Savard law talks about the B field produced by a steady current I. It is written

[tex]\vec{B}(\vec{r}) = \frac{\mu_0}{4\pi}I \int\frac{d\vec{r}'\times (\vec{r}-\vec{r}')}{|\vec{r}-\vec{r}'|^3}[/tex]

Earlier on the same page, he also say that a moving point charge cannot be considered a steady current. Hence Biot-Savard law and moving charges have nothing to together. At least that'd what I concluded.
Aug30-05, 10:40 PM   #9
 
but current is ultimately made up of moving charges only .so why biot savarts law cant be applied for moving charge. can u say an electron moving in orbit doesnt generate magnetic field at the centre?
Aug31-05, 02:55 AM   #10
 
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Quote by shahidkhilji
but current is ultimately made up of moving charges only .so why biot savarts law cant be applied for moving charge.?
Good question. So it is true then? A moving charge produces a circular magnetic field around itself.. which is given by

[tex]\vec{B}(\vec{r}) = \frac{mu_0}{4\pi} \frac{q( \vec{v} \times (\vec{r}-\vec{r}'))}{|\vec{r}-\vec{r}'|^3} [/tex]

where [itex]\vec{r}'[/itex] denotes the position of the charge. That's pretty darn cool, and a real shame (on David J. and my teacher!) that I've been through an E&M class and didn't know of this elementary fact.

I also suppose that this formula is only an approximation for then v << c, right? The real formula probably takes the delay into account.


Quote by shahidkhilji
can u say an electron moving in orbit doesnt generate magnetic field at the centre?
Now I am forced to.
Aug31-05, 01:18 PM   #11
 
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Quote by mukundpa
magnetic field at q1 due to q2 is not constant. did you accounted the electric field at q1 due to changing magnetic field of q2 at q1?
Do you mean because of Maxwell's equation [itex]\vec{\nabla}\times \vec{E} = -\partial\vec{B} / \partial t[/itex]? Is there a way to calculate exactly the electric field created by this change of B (i.e. w/o assuming that newton's third holds)?
Aug31-05, 08:45 PM   #12
 
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Hold on! IN THE LAB FRAME, the B-field at any LOCATION
on the plane perpendicular to v_2 is reaching a maximum,
so dB/dt (at the place Q1 is at the time) is zero.
(Since we're not told actual distances nor velocities,
I wont include the propagation delay explicitly ;
the line of dB/dt = 0 is really along tan(theta) = v/c.
The E-field and B-field disturbances propagate at c, so
the 2 subscripts on each F are displaced by t=d/c.)

Q1 is moving into a region of stronger B,
but if you want to use dB/dt for this, you need to
find new E and B, the calculation in Q1 REST FRAME.

What Shahid didn't "get" was that you really NEED
to include the dE/dt (displacement current) term,
since this is not a static situation.
At the location of Q1, this is zero, while at Q2 it's not.
Aug31-05, 11:40 PM   #13
 
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Damn this is complicated. Maybe I'll check this thread out again after my E+M waves class (starting... tomorrow! ).
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