good morning sir😊 @JohnRennie

@user8718165 morning :-)

@JohnRennie Sir I've a question 'bout factorials

@user8718165 hi, what's the question?

@JohnRennie sir Its only defined for whole numbers...right

@JohnRennie okay sir :-) but what does that actually mean...like number of ways of arranging $1/2$ of the objects...?

@user8718165 it doesn't have a physical meaning as far as I know.

The gamma function is a function that gives the same value as the factorial for integers but it defined for non-integers as well.

But that doesn't mean its values for non-integers are related to anything physical like probabilities.

Mathematicians like that sort of thing :-)

@JohnRennie You and I dont XD

@JohnRennie okay sir...got it:-)

hello:-) @JohnRennie

@user8718165 hi

@JohnRennie hi.

@Nobodyrecognizeable hi

I wanted a proof that a particle with velocity perpendicular to a constant magnetic field always moves in a circle.

@JohnRennie ^^ do you have any reference. And I'm referring to mathematical proof.

The proof is imply that for this particle the acceleration has a constant magnitude and is always perpendicular to the velocity, and in circular motion the acceleration has a constant magnitude and is perpendicular to the velocity. Ergo the motion is circular.

@JohnRennie OK.

@JohnRennie but do you have any referencec?

I'm not sure what reference you need. You can easily prove for yourself that in circular motion the acceleration is always normal to the velocity and of constant magnitude.

@JohnRennie if I wanted to set up the equation of motion then?

@Nobodyrecognizeable For circular motion $x(t) = A\cos\omega t$ and $y(t) = A\sin\omega t$ where $A$ is the radius. Yes?

How would you prove that?

@JohnRennie sure.

To get the velocity differentiate the position wrt time. So the velocity is:

$$ \mathbf v = (-\omega A \sin\omega t, \omega A \cos\omega t) $$

@JohnRennie yep.

And differentiate again to get the acceleration:

$$ \mathbf a = (-\omega^2 A\cos\omega t, -\omega^2 A \sin\omega t) $$

@JohnRennie - $\omega ^2 $ x both.

@JohnRennie force is indeed q(vxB)

Yes, you need to know the equation for the Lorentz force.

@JohnRennie presumably B is in x or y to minimise the calculation.

For a circle in the $xy$ plane the magnetic field would be normal to the plane i.e. along the $z$ axis.

@JohnRennie all right. I'll get different component of forces to set up the equation of motion.

@JohnRennie the force is thereby $(\omega ABcos\omega t, \omega ABsin\omega t) $

Thinking of B=(0, 0,B)

@JohnRennie hey are you here?

Can you clarify what you did? You started from $\mathbf x = (A\cos\omega t, A\sin\omega t)$ and computed the velocity then used the Lorentz force equation to calculate the force?

@JohnRennie yup.

OK, I'd have to do the calculation to check, but you've shown the force is in the same direction as the acceleration calculated from the circular motion.

@JohnRennie yep.

So comparing the two equations you get $\omega = qB/m$

@JohnRennie which is a constant. All right.

@JohnRennie so that's it. Are we done?

Really a great proof. Thanks professor for the worthy help; whereas I have the exam tomorrow. Have a nice day professor.

@Nobodyrecognizeable well you've shown that the acceleration due to the Lorentz force has the same magnitude and direction as the centripetal acceleration.

@JohnRennie great point though. Never thought of it. Anyway have a nice day professor. Goodbye.