Magnitude of the field used for electron

AI Thread Summary
The discussion revolves around calculating the magnetic field required for an electron to follow a circular path with the same radius as a proton in a magnetic field of 0.5 T. The relevant equation used is r = mv/qB, where the mass of the proton is significantly larger than that of the electron. The user initially calculated the radius for the proton and then applied the same equation to find the magnetic field for the electron, arriving at a value of 2.7 x 10^-4 T. It was noted that the charge of the proton is opposite to that of the electron, which affects the direction of the magnetic field. The conversation emphasizes the importance of correctly applying the principles of electromagnetism and the relationship between mass, charge, and magnetic fields.
flynostrich08
Messages
20
Reaction score
0

Homework Statement



a proton is projected perpendicularly into a magnetic field that has a magnitude of .5 T. the field is then adjusted so that an electron will follow a circular path of the same radius when it is projected perpendicularly into the field with the same velocity that the proton had. What is the magnitude of the field used for the electron?

Homework Equations


r=mv/qB


The Attempt at a Solution


i basically used the above equation to solve for r but left v=1 since it is the same in both then used B=mv/qr and plugged in r and realized i just did the same thing so can someone please help explain what i have to do in detail since i am lost. Thanks for any help
 
Physics news on Phys.org
It should help you knowing that the proton mass is 1836 times bigger than the electron mass (mp=1836me)
 
thanks i was looking at my book wrong so i ended up solving for r of the proton then took that and solved for B of the electron using the same equation and got 2.7x10^-4. did anyone else get that?
 
flynostrich08 said:
thanks i was looking at my book wrong so i ended up solving for r of the proton then took that and solved for B of the electron using the same equation and got 2.7x10^-4. did anyone else get that?

I got the same result but it has to be negative because of the charges (qp=-qe). That result is for the same radius (distance from the center) but in the opposite direction.
 
thanks for pointing that appreciate the help
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top