Accelerated from rest through a potential = difference to a In a linear accclerator; protons are proton beam enters a region of uniform magnetic ~speed of fapproxim...

$\bullet$ A beam of protons is accelerated through a potential dif- ference of 0.745 $\mathrm{kV}$ and then enters a uniform magnetic field traveling perpendicular to the field. (a) What magnitude of field is needed to bend these protons in a circular arc of diameter 1.75 $\mathrm{m} ?$ (b) What magnetic field would be needed to produce a path with the same diameter if the particles were electrons having the same speed as the protons?

High. In the given problem, potential difference used to accelerate the beam of protons means positive. Judge is given as we is equal to 0.745 KT. All we can say this is 745 vault in the first part of the problem diameter of the circular arc the proton makes in the magnetic field. It's given us 1.75 centimetres. So the radius of that circular arc. Yeah, yeah, We'll be able to. Half of the diameter means 1.75 Centimetre divided by two means this is 0.875 centimetre. And here this radius is already in NATO. Not in centimeter. No. Using the expression for the radius of the charged particle when it moves in a magnetic field. That is given as the product of mass with the velocity means the linear momentum divided by the magnet field and charge. And there is a relation between the linear momentum and kinetic energy. Linear momentum is equal to twice of square root of twice of the product of mass with and energy kinetic energy. And that kinetic energy assumed by the proton will be given by product of the charge with the potential difference applied to MTV by the cube. So plugging in holland values here, but before that, exchanging the be magnetically with the are because we have to find a magnitude of magnetic fields. So this magnetic field will be given by The square root of two M. Q. is the charge of a proton divided by radius into cuba. Again, the charge for a proton finally, now plugging in all known values. This is two times of the charge or the proton which is 1.67 into 10 for -27 kg multiplied by the charge of proton, which is 1.6 into 10 days for -19. Coolum multiplied by the potential which is 745 World. Then divided by radios, which is 0.875. Multiplied by the charge. Again, it's 1.16 to 10 for -19. And here it will come out to be the square root of 3,981 point to it into 10 for -46. Divided by 0.875 Into 1.6 Into a 10 for -19. And finally it becomes 63.1 into 10 for -23, divided by 1.4 in 2, 10 for -19. And finally, this magnetic field comes out to be 4.51 in 210 Dish par -3. Tesla, which is the answer for the first part of this problem. No. In the second part of the problem, the proton is being replaced with electron, so on replacing brought down by electron. The only change will be in the mass is the mass of the electron is 9.1 Into 10 for -31 kg. So putting all these values here again in the same medication, Be comes out to be Square root of two In 2 9.1 into 10 for -31 Into 1.6 Into. 10 day 4 -19 Into potential. Which is same. 700 45. World divided by zero point its own fight. The radio's into charge 1.6 into 10 for -19 cola. So finally, this magnetic field here comes out to be the square root of 21694.4 into 10 days. For -50, Divided by 1.4 In 2, 10 days, 1 -19. And finally, this is 1.05 into 10 to the power minus four. That's love. Answer for that 2nd part of the problem. Thank you.

Question. We have a proton which is accelerated from rest by a potential difference. Delta V. Equals to 400 world. And the proton then enters a magnetic field and follows a circular path of radius small R equals to 20.0 centimeters. That is 0.2 m. And we have to determine the magnitude of magnetic field B. Okay, so we can write from the kinetic energy or energy conservation. So delta kate, this will be equals two delta. You cannot take energy and this is potential energy and kinetic energy K is given by the multiplier by we where we is the potential difference. So we can call it at delta P. And this will be equals two. one x 2. M. We square where we is the speed of the partick. Okay, so from here still we equals two E delta. We they were very small and the whole under route. Ok. And we know that migrant it will be is given by MB by E manipulated by our so substituting value of we hear. So um by E are manipulated by we which is equals two under root of E. Delta. We divided by mass small M. Okay, so from here, after rearranging we will get that one by our And the root of two. M. Teldta we develop a charge. So now substituting values in this equation. So we will get that be this will be equal to one by our which is 0.2. And under route To manipulate by mass on the proton, which is 1.67 into 10. To the power -27 multiplied by potential difference which is 400 gold divided by charge on the electron which is electron or proton. So 1.60 to manipulate and to the power minus 19. Cool. Um so from here, after solving we will get magnitude of magnetic field B equals to 1.44 multiplied by 10 to the power minus two. Tesla. Okay, So this is the answer for this question, and this is the required magnitude of magnetic field. Okay?

High in the given problem, vertically, upward electric field Is created with the help of two plates as the field is vertically upwards. So the lower plate should be positively charged and upper plate should be negatively charged. The direction of electric field then becomes upward. Now the proton being, which is allowed to enter here in this electric field horizontally moving with a velocity V. So this proton beam is accelerated Bye, A potential difference of We is equal to 3.0 kilovolts. Or we can say this is 3000 votes. The gap between these two plates is given as D is equal to and Centimetre. Or we can say this is 0.10 meter. The potential difference of light between the place Is given as let it be videsh and that is equal to 250 gold. Now, in the first part of the problem, we have to find the speech of this proton before it enters into this electric field, for which we use the concept and energy kinetic energy gained by the electron. Uh sorry, proton means half and be square will be equal to its electrostatic potential energy that is given by the product of charge over the proton. With the potential difference which is used to energize. It means half family squared equals to ian to be. So. This expression gives us the expression for the velocity of the proton. This is the square root of two E. V. Divided by M. Where M. Is the mass of a proton. So, plugging in all non values For the charge over proton, this is 1.6 into 10 days par -19 cola. For the potential through which it is being energized 3000 world, divided by mass of a proton, Which is 1.67 to 10 days to power -19. Sorry, this is 1.67 to 10 days par -27 kilogram. So finally, this speed achieved by the electron. Sorry, proton Either squirreled of 5748 point five into tennis part eight, or we can say this is 7.6 into 10 days for five meter for second. Which is the answer for the first part of this problem? No. In the 2nd part of the problem. To pass this proton being under flecked ID, what should be the magnitude and direction of the magnetic field applied? We have to find it so to pass the proton being un deflected. As this proton beam will be experiencing an electrostatic force F. E. In upward direction towards negativity. Some magnetic force should be acted on it in downward direction. F B should be vertically downward. To pass the program, demand deflected magnetic Lawrence force in downward direction should be equal to electrostatic force in upward direction means E. G into B is equal to E into E. Where this capital is the electric food and has electrical is given as the potential gradients. So this is potential difference applied across the planet's videsh divided by gap between them, so canceling discharge over the proton. Finally, expression for magnetic failure comes out to be The dash by small with the speed of proton in today. So, plugging in non values 250 volt for the potential difference applied between the players divided by we the speed of the proton which we have obtained in the first part of the problem 7.67 to 10 days apart, five m per second. Multiplied by the gap between them, which is 0.10 m. So finally, this magnetic field which should be applied, comes out to be 32.4 into tended bar minus for Tesla. Or we can say this is 3.3 millie Tesla approximately answer for the magnitude of this magnetic field? No. For its direction. As this magnetic Lawrence force should act were typically don't work to balance electrostatic force. We have already discussed in the fever. So using the expression for magnetic Lawrence force in vector form F is equal to Q. Into we cross the. So using right hand screw rule, the direction of magnetic field should be outward perpendicular to the plane of paper. So this becomes a complete answer for the second part of this problem. Now, finally, In the 3rd part of the problem, if the magnetic will become small more than that which we have found in the second part of the problem, then the magnetic force acting on it will become more and that is in downward direction. So under that magnetic force, the proton, the proton beam will meat a circular part in the plane of paper within the electric field. And here it is, the answer for the third and the last part of this problem. Thank you.

Solving party of this problem. So here I can write the formula edge F C. Is equal to M B squared by us. Which on simplification I can write Q. B. B. Which is equal to F. C. Is equal to M. B. Square By our simplifying it further, I can devalue A B is equal to M B by Q R. Which is equal to 1.67 multiplication and 10 to depart minus 27 kg multiplication and 1.38 multiplication, 10 to the past six m purse again by 1.6 multiplication and 10 to the power minus 19 column multiplication and 0.5 m on simply because an idea developed be at 0.3 D. Now solving part B. So in part we I can like the value of L is equal to two pi R by food which is equal to two pi multiplication. 0.5 m. Buy food Which is equal to 0.785 m. Now time required is equal to L by P, which is equal to 0.785 by 1.38 multiplication and 10 to the politics which is equal to 5.7 multiplication and 10 to the call minus seven second. Now For party I can like a physical two QBB. On simplifying it further, I can write the value of F is equal to 1.6 multiplication, 10 to the power minus 19. Coolum multiplication, 1.38 multiplication, 10 to the past six m per second, multiplication, 0.3 D. So on simplification idea the value of effort 6.6 multiplication and tend to the par minus between newton as our final answer for party.