As an advanced lab project, you are to develop a lab similar...

As an advanced lab project, you are to develop a lab similar to the original e/m experiment. You find a tube (shown below) which has a filament heated by an AC source, a cathode, anode, collimator, horizontal deflecting plates and a fluorescent screen. You have access to several power supplies and a "U" shaped magnet that produces a 1.25 × 10?4 T magnetic field. You connect a power supply to the cathode and anode (as shown) and adjust it to 45.0 V. The separation of the deflector plates is 5.00 mm and they are 2.50 cm in length. power supply Va + S d L + ? N filament cathode collimator anode i (a) Determine the speed (in m/s) at which electrons enter the region of the crossed electric and magnetic field. m/s (b) Determine the electric potential (in V) which must be applied to the deflecting plates in order for the electrons to pass straight through the region of the crossed electric and magnetic fields. V (c) If the magnetic field is removed and the electric field kept the same, determine the vertical deflection of the electrons (in mm) as they exit the region of the electric field. mm

Transcript

0:00 Hello everyone.

00:01 So in the question we had to give a nail in that project in which a tube which heating by an pc source and there is two magnets around the probe which produce the magnetic field.

00:16 So an electron is ejected in magnetic and electric field to do the adjusted potential apply here and this is the partition between the detective field and is the length here.

00:29 Okay, so what we have to do is the whole we have to find out here the speed at which the electron enter the region and crossing the electric and magnetic field.

00:40 Okay, so to find the speed first of all we have to given here potential that is 45 gold so from a law of polarization of from contervation of energy what we do here? by conversion of energy, we do that half n -2 square is equal to qb.

01:10 Okay, according to the last generation of energy, the magnetic energy that is converted is equal to the core supplied by the n equal.

01:19 This is equal to qb into b.

01:21 Okay.

01:22 So from here, if you find out the value of u, then u b will be equal to count b, b, multiplied by 2, n -20, this term, by s.

01:31 And the square root of the stem because this square comes towards the cell and the square root of the term okay so this will be equal to under the road multiply by 2 the value over 2 is here 1 .6 multiply by 10 plus to the power minus 9000.

01:51 We're going into charge of the vector and v is that is equal to us that is 45 okay and divided this term by nm n is mass of director that is 9 .1 multiply by 10 to the power minus 30 okay so by all of these terms we go that the value of u hr 3 .978 multiplied by 10 to the power 6 meter per second so this is the value of velocity here okay, so in the next part i'll see that we had to find out the value of v.

02:33 So what we have to do in here to find out the value of to find out the value of electric potential.

02:44 Okay, so what do we do here? so let we hit the protection applied for this mirror.

02:53 Let we be the black.

02:55 The potential applied.

03:04 Okay, so what happened here? here e is equal to b divided by d.

03:10 Okay, and if we multiply this term, it is more e, category, this will be the value of force, and this will be e ub according to the state line.

03:25 Okay, in the state line part, delta 4 supply is equal to e, u and b.

03:34 So from here we find out the value of e so e will be equal to u and b okay so this will be equal to from here if we find out the value of b so b will be equal to e multiply by e so b will be equal to here e that means u into v and multiply by b so the value of u we have already teluguity that is 3 .978 multiplied by 10 to the power 6.

04:07 Okay and this is given in the question that is 1 .125 multiplied by 10 to the power minus 4 and then the next multiply is combined and value is 5 multiplied by 10 to the power minus 8 why managed to do not be able to see if the only age demurement.

04:30 Okay? so, um, by so you see the value of bd2 .4862.

04:39 So this will be the potential in the electric field.

04:42 Okay, so the value of electric field in the c part we had to find out here time breaker.

04:49 Okay, so much of all we find out the value of electric field that is equal to be divided by d and this will be equal to be divided by d...

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