Conversation started Feb 22, 2020 at 6:57.
Guru Vishnu
Guru Vishnu
Feb 22, 2020 06:57
@JohnRennie: Hi sir. Good morning :-)
John Rennie
John Rennie
@GuruVishnu hi :-)
Guru Vishnu
Guru Vishnu
@JohnRennie Are you free now sir? I need some hints to solve a problem which I analysed it only qualitatively.
John Rennie
John Rennie
@GuruVishnu I'm working, but only for a few minutes. Post the question and I'll have a look as soon as I'm free.
Guru Vishnu
Guru Vishnu
Ok sir. This is the question:
A long straight wire carries a current $i$. A particle having a positive charge $q$ and mass $m$, kept at a distance $x_0$ from the wire is projected towards it with a speed $v$. Find the minimum separation between the wire and the particle.
Thank you.
John Rennie
John Rennie
Feb 22, 2020 07:16
@GuruVishnu that looks hard to me ...
Guru Vishnu
Guru Vishnu
@JohnRennie Hmm... The situation of non-uniform magnetic field with varying distance makes the situation hard. Is there anyway I can determine its path? Anything as simple as virtual work?
One thing I can say for sure is: The speed remains constant. I know how the magnitude of force varies. But I'm wondering on how to find its direction at different points of time.
John Rennie
John Rennie
user image
The motion would look like this
Guru Vishnu
Guru Vishnu
@JohnRennie Is that symmetrical about the horizontal axis? I think yes.
John Rennie
John Rennie
Yes, it would be.
Guru Vishnu
Guru Vishnu
@JohnRennie So, will it be an elliptical path?
John Rennie
John Rennie
Feb 22, 2020 07:26
But I can't see how to calculate the trajectory. I doubt it would be elliptical.
Guru Vishnu
Guru Vishnu
If we consider the motion beyond the right edge of the image?
John Rennie
John Rennie
This is surely not a JEE question?
Guru Vishnu
Guru Vishnu
@JohnRennie Ok sir. But I suspect it would be much different than planetary motion as speed here remains constant, whereas in planetary motion speed is maximum at perigee and minimum at apogee.
@JohnRennie I think this is not a JEE question. But it seems I could answer it based on my understanding, although I couldn't figure a way out.
John Rennie
John Rennie
The differential equation is easy to write down. Solving it is the problem!
Guru Vishnu
Guru Vishnu
@JohnRennie Fine. Can you just give a brief idea on how to approach it?
At least I'd like to have a good idea of it rather than solving it.
John Rennie
John Rennie
Feb 22, 2020 07:37
Take $x$ to be the distance from the wire and $y$ the vertical position. Then we can write the velocity of the particle as the vector $\mathbf v = (v_x, v_y)$
Guru Vishnu
Guru Vishnu
Ok sir. And $v^2_x+v^2_y=v^2=\text{constant}$
John Rennie
John Rennie
The force on the charge is the Lorentz force $F = q \mathbf v \times \mathbf B$ where $\mathbf B = (0, 0, \mu_0 I/2\pi x)$
Guru Vishnu
Guru Vishnu
Ok sir.
John Rennie
John Rennie
If you take the cross product you'll fund the force is always in the plane of the page so you can write it as $F = (f_x, f_y)$. Then the acceleration is $\mathbf a = (a_x, a_y)$
And you'll end up with a differential equation $\mathbf a = d\mathbf v/dt = f(\mathbf v, x)$
Guru Vishnu
Guru Vishnu
Ok sir. It seems we're just decomposing two dimensional motion into two single dimensional motion along orthogonal directions to make the situation easier like we used to do for projectile motion.
John Rennie
John Rennie
Feb 22, 2020 07:41
Good luck solving that!
Guru Vishnu
Guru Vishnu
@JohnRennie :-)
@JohnRennie: Is there anyway I could find "Magnetic potential energy" at some distance from a current carrying wire? Is it similar to other such forces like gravity and electrostatic?
I think we may be able to apply law of conservation of energy?
John Rennie
John Rennie
I don't know of any such method.
Guru Vishnu
Guru Vishnu
Ok sir. Thank you.
¯\_(ツ)_/¯
I think it's a challenging task.
John Rennie
John Rennie
$$ \frac{dv_x}{dt} = \frac{\mu_o I q}{2\pi m} v_y / x$$
$$ \frac{dv_y}{dt} = -\frac{\mu_o I q}{2\pi m} v_x / x$$
That's what I get for the equations of motion.
Guru Vishnu
Guru Vishnu
I got $v/x$ instead of $vx$ sir.
For both $x$ and $y$ components.
John Rennie
John Rennie
Feb 22, 2020 07:53
Oops, yes
Guru Vishnu
Guru Vishnu
I'm planning to do that after some time.
So I might get some better ideas.
 
Conversation ended Feb 22, 2020 at 8:00.