(b) Suppose a particle moves under an attractive force...

(b) Suppose a particle moves under an attractive force (i.e., a force directed towards the origin) with magnitude per unit mass $\frac{A}{r^2} + \frac{B}{r^3}$ where $A > 0$ and $B > 0$. (i) Show that the general solution for $u(\theta)$ in the case $B < h^2$ is $u = \frac{A}{h^2\omega^2} + \alpha \cos(\omega\theta) + \beta \sin(\omega\theta)$ where $\alpha$ and $\beta$ are constants of integration and you should give an expression for $\omega^2$. (ii) Find the corresponding general solution for $u(\theta)$ in the case $B > h^2$. 2. Return to the model in Question 1(b) and suppose that initially the particle is located at $r = b$, $\theta = 0$ and is moving in the tangential direction with speed $V$ such that $A = bV^2$ and $B = b^2V^2/2$. (a) Determine $h$ and the initial values of $u$ and $\frac{du}{d\theta}$. (b) Find the particle's trajectory $r(\theta)$. (c) Show that the orbit is bounded. (d) Explain whether the orbit ever repeats itself. If it does, find the period.

Transcript

00:01 Hello everyone so let's start with a question a charge particle having a mass of 4 .4 multiplied 10 to minus 8 kg moves in with a constant velocity in the wide direction enters a region constantly containing a constant magnetic field b equal to 2 .27 tesla b equal to 2 .2 tesla aligned with a positive xx as you can see in a figure okay it looked like some kind of this okay here these are our y -axis and here and these are xxies so the magnetic field is towards the positive that means z xx is like this in this direction if i consider this as a positive xx so it will so our magnetic field is in this direction now the particles enter the region at the x y means suppose the particles is enter the region okay suppose a charge particle which is enter a region okay this our charge particle so it enters the region of that is 0 .73 and leaves the region at the with the that is here as you can see that is point 73.

01:16 At the time of means that gives the time at 875 microsecond after it entered the regions.

01:23 So means as you can see the coordinates we have given in the question that the particle is enter when when it enters there with the coordinates of points on 8 meter and it leaves with a coordinate of 0 .78 meter okay with the time period of 875 microsecond so what will be the speed of a particle when it enters the region also we have to find the horizontal component of force acting on a particle at the time when the at the time period of 291 .7 microsecond after it entered the region containing the magnetic field also we have to find the vertical component of force acting on a particle at time of 291 .7 microsecond after it enters the region containing the magnetic field.

02:06 Also we have to find the what is the charge of a particle.

02:10 We don't know the charge of particles so we have to find the q value.

02:14 And if the velocity of the incident is charge particle is double, then how would be our magnetic field have to change? if the velocity of a particle is doubled, then what will be the magnetic field will be.

02:28 So these are the given option we have to select on the given option.

02:31 So let's start with the solution.

02:33 Here, if we do analysis of this figure, we get that as the particle moves in the magnetic field, where it is which is perpendicular to the speed, okay, means if the particle is moving in this direction or means it's moving into this direction and enters and leaving to this direction.

02:57 So our magnetic field is perpendicular to the speed and it takes a circular path of a radius which which is at the center if i draw the exact movement of the particle in magnetic field it look like this that is so the trajectory of the particle will look like this that it enters it enters in this direction means in the coordinates of points on it and leaves the in this direction that is point seven eight okay so it will be suppose if the if the particle is here so what it will be okay so magnet means centipital force is in this direction that is mv square by r and magnetic force will be in this direction okay and here our velocity of a particle will be means perpendicular to the magnetic force as as you know so if the particle moves in the magnetic field which is perpendicular to the speed then it takes the circular path of radius okay so this this is our radius of circular path okay like of same distance that is point seven eight which lies in the center of the region okay so as the particle moves from if the particles moves from point seven eight comma to zero comma point seven eight meter so our velocity will be what velocity is equal to this distance upon time so here the distance is what pi by 2 if i want to talk about in like radian.

04:39 So our distance is what pi by 2, pi r by 2, sorry.

04:44 Okay, it takes a first quadrant of a circle.

04:47 So our velocity will be what? pi r divided by 2 multiply by time.

04:51 Here the time is given that is 875 microsecond and we know the radius that is 0 .78 meter.

04:59 So after putting the value in this equation, that is, pi multiply by 0 .78 divided by 2 multiplied by 875 microsecond multiply by 10 to minus 6 so velocity will be what 1 .4 multiply by 10 to 3 meter per second okay so these are velocity okay this this is our answer for our first question now for the second we know that if the trajectory of a charged particle is perpendicular to the magnetic field then the force acting on the charge particle will be what that is f q is what q q v cross b okay if i want to write in the magnitude so means the velocity having a direction the velocity having a direction in y direction and the magnetic field is a positive k direction that is mentioned in the question so fq i can write that is q that is v j cap cross b k k k k k k k k k k and also we know that j cross k is gives us i cross okay so f q is given by that is q magnitude of v magnitude of b i cap okay so the our force having a direction means the force will be act in x direction at center and it always perpendicular to the to it okay so so our force we have no now for the time taken by particle to trace quarter circle is how much 875 microsecond.

06:50 So here as you can see in the diagram means it trace the quarter circle that is this is our quarter circle and it required time of 875 microsecond.

07:00 So we have to find the force in x direction when the time is 291.

07:05 Micro 7.

07:06 Okay.

07:08 So i can write that is i can write so i can write like it take 875 microsecond to trace quarter that is 90 degree so 291.

07:29 So 291 .7 microsecond means how much required the circle part okay so we have to find this so if i consider this our theta so theta equal to what that is 90 multiply by 291 .7 divided by 875 okay after solving this we get our atrox 30 degree so for the time to 91 .7 the particle trace 30 degree angle means it covered 30 degree so i can write that is if i draw a x and y component so this is our trajectory so our force f is and this is our velocity and so this will be this is our theta so it will be what f x cos theta and this will be f y sine theta okay so i can write that is f x is equal to q b b cos 30 degree f y uvb sign 30 degree.

08:40 So we know the value of velocity, magnetic force, magnetic field, but we don't know the charge.

08:46 So first we have to find the charge value...

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