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At a particular instant; an electron is traveling north with a kinetic energy of 7.8keV. Earth's magnetic field has a horizontal component of 1.8x10-ST west an...

Question

At a particular instant; an electron is traveling north with a kinetic energy of 7.8keV. Earth's magnetic field has a horizontal component of 1.8x10-ST west and vertical component of 5.0x10-5 south: (a)What is the moving direction of the electron? (draw it on the paper) (b) What is the radius of curvature of the path?

At a particular instant; an electron is traveling north with a kinetic energy of 7.8keV. Earth's magnetic field has a horizontal component of 1.8x10-ST west and vertical component of 5.0x10-5 south: (a)What is the moving direction of the electron? (draw it on the paper) (b) What is the radius of curvature of the path?



Answers

At a particular instant, an electron is traveling west to east with a kinetic energy of $10 \mathrm{keV}$. Earth's magnetic field has a horizontal component of $1.8 \times 10^{-5} \mathrm{T}$ north and a vertical component of $5.0 \times 10^{-5} \mathrm{T}$ down. (a) What is the path of the electron? (b) What is the radius of curvature of the path?

Mhm. This problem covers the concept of the magnetic force on a movie charge, the electron, which is moving uh along not direction, so the velocity of the electron is along the north direction and the fourth given on the electron is longer northeast direction, that is the longest production. Okay, now we know they're the force equals the magnitude of the charge into the velocity times the magnetic field into science teacher. And the direction of the force due to the magnetic field on the charged particle is always perpendicular to its velocity, and the force is always perpendicular to the magnetic field direction. And in the problem, the angle between the velocity and the forces 45 degree. So we can say the angle, but when velocity and the magnetic force on electron is not equal to 90 degree, hence the situation, it's not possible. And the correct option is option E.

Have an electron having charged negative E. And it is traveling horizontally from northwest towards southeast regions. We can draw that its velocity is pointing in this direction. Okay? And uh enters in a region of space where earth magnetic field is directed horizontally towards the north. So Earth magnetic field is pointing in this direction. So this is the vector and this is velocity vector. Okay? And this is the angle between them supposed to. So now the magnetic force F vector on a charged particle is given by Q velocity vector. Cross diverted. Okay. And charges here negative. So we can write that minus E. And velocity vector cross the vector. So from the right and rule we can say that the direction of the magnetic force on the negative charge particle which is minus E, electron, it will come out to be involved. Okay. And what that is cross? So the direction of the force it will be in word or cross inward into the plane of paper. So this will be the direction of magnetic force on the charged particle when it enters in a magnetic field which is pointing in the north direction. Okay.

So when we project the charge in magnetic field at some angle theta then if we take the two components here, one component will become because to turn over the component becomes the scientific and since the we sent it is perpendicular to be because of recent it eight will perform circular motion and because of the cost retired we're moving forward direction. So because it will not change. So in order to find the radius of so the result is it will follow the helical path. So radius we know the radius of medical part. He's given as R is equal to M in two perpendicular component of the perpendicular to the magnetic field. So this is the perpendicular component. So I will become equal to um we scientists to what can be now if we substitute the values here, The value of them is nine. You just Electrons have mass effect on this 19-23 power -31. The value of we is given four into 23 power five. The value of scientific 30 to 45 degree And signed 35° 0.6. Okay 7 35. 0.6. The charge on electron is 1.16 to 10 days to power -19. And the value of magnetic field is 0.04. So on solving the Reedus council to be three point 37 into 10 days to £5 minus five m. No let us solve further for people that is how much it will move in one complete cycle. So this was the airport. So for be part we need to find a pitch of political party so which we can write the viral component of B into time period. My little component is we cost data in time period is to buy em over QB so PHP we can substitute the values here. Let us arrange first it will become too I am we cause tater work you be now let us shoot the values here. So to the value of Pi is 3.14. Moss is 19- 23. Power -31 Velocities four into 10 days to power five. The Value of Course 35 0.79. The value of charges 1.67- 10 days to power -19. Yeah And the value of my addict food is 0.0 for Tesla On solving it comes out to be 310 into 10 to the power -7. And on for this all the value of P comes out to be 3.1 in took into the power minus five m. This is the answer. Thank you.

We have an electron which is moving with speed V equals to six point double zero flour bait. And to the power seven m per second perpendicular to the earth magnetic field. So magnetic field be it will be equals to uh 0.25 double zero. But played by 10 to the power minus four Tesla. And we have to determine the radius of uh electron circular path. So radius of electron circular path, it is given by Mv by QB. Were Q. Is the charge on electron magnitude only. And this is the mass on electron. Okay, Okay, so mass of electron is nine point 109 Multiplied by 10 to the power minus 31 kg. Molecular based will be which is 6 to 10 to the power seven m per second. They were by magnitude of charge on electron which is 1.60 to molecular by 10 to the power minus 19, particular by magnetic field which is 50.5 W. Particular by 10 to the power minus four Tesla. Okay, so from here, after solving, we get radius of electron, circular path are equal to 6.82 m. Okay, so this is the answer for this question. Okay, this is the radius of the circular path for the electrons. Okay.


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