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05:00 - 17:0017:00 - 18:00

5:29 AM
@AvyanshKatiyar hi. I'm generally only here in the morning. From about 11:00 a.m. Indian time.
@JohnRennie Good Morning
@YouKnowMe hi :-)
@JohnRennie Two wires carrying current in the same direction attract each other, and they repel if the currents are opposite in direction. I am not able to find this by Left Hand Rule.
Let's draw a diagram:
I've drawn two wires with the current going into the screen.
And the circle is a field line generated by the left wire. I've marked the direction of the field.
@YouKnowMe OK so far?
5:46 AM
@JohnRennie Yes
So consider the Lorentz force on the electrons in the right wire. Their velocity is into the screen because the current is into the screen, and the field at the wire is downwards. The force is F = ev x B. Yes?
@JohnRennie Yes
And when we take the cross product the force is to the left
No, wait, that's wrong. I forgot the electron sign.
@JohnRennie Why there will be a sign on electron?
The charge on an electron is -e
5:56 AM
Doesn't e mean $−1.602176634×10^{−19}$ C?
Hmm, yes, I'm also getting a repulsion ...
That's weird
I would normally use $e$ to mean $+1.602 etc$
Let me go away and think about it
@JohnRennie Ok
Hello sir
@PrateekMourya hi :-)
6:11 AM
Sir can we say that if kinetic energy of a body is constant and if we choose axis of rotation them MOI is measure of how much omega it will get about that axis?
Yes, you could look at it that way.
That will help me build and intuition for it
I'm not sure how helpful this will be, but we can try it.
From that i can derive torque and angular momentum
Suppose the particle is moving at a velocity v and we take a point a dstance r from it as our axis, then the angular velocity is ω = v/r. Yes?
6:14 AM
In the chapter com i accepted definition of com because archemides experiments proved it
The MOI about the axis is I = mr²
Yes as far as we accept
But I'm not sure where this is going ...
@PrateekMourya suppose we calculate the angular KE for our particle. Using the definitions of ω and I we get:
KE = ½ I ω² = ½ (mr²) (v/r)²
OK so far?
6:25 AM
And that gives KE = ½ m v²
And you should recognise tht equation as just the KE of a particle moving at speed v
So specifying ω and I this way gives us the same KE for the particle as the usual linear KE.
6:26 AM
So relations between angular and limear quantities
7:03 AM
@JohnRennie Sir, Why are you calculating magnetic force on electrons instead of positive charges?
@YouKnowMe because it's the electrons that are moving in the wire
10 hours later…
4:38 PM
@JohnRennie You must be busy right now.
@YouKnowMe eating lunch ...
05:00 - 17:0017:00 - 18:00

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