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
Yes

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
Ok

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 ...

5:07 PM
@YouKnowMe hi, I have finished lunch if you're still around.

5:24 PM
@JohnRennie Draw the line of force(indicating field direction) of the magnetic field through and around (a) a single loop of wire carrying electric current, and (b) a solenoid carrying electric current.
What does "Lines of force" mean?

A line of force shows the direction of the force the field exerts on a charge.
(For a magnetic field this would have to be a hypothetical charge since there are no magnetic monopoles)
So suppose you are holding a charge in a field and you move the charge slowly so that you are always moving it in the direction of the force that the field exerts on it. The motion would trace out a line of force.
You may know them as field lines.

@JohnRennie Magnetic field lines?

Yes, or of course electric field lines if the field is electric.

Thanks :-)