Magnetic Force to balance out Electric Force

AI Thread Summary
An electron accelerated from rest through a 2.0 kV potential difference enters a gap between parallel plates with a 130 V potential difference and 20.0 mm separation. To maintain a straight trajectory, the magnetic force must balance the electric force, leading to the equation vB = E. The velocity of the electron is derived from the work-energy theorem, resulting in B = (V2/d)√(m/(2qV1)). The calculated magnetic field values yielded discrepancies, with one calculation resulting in 2.45 x 10^-4 T and another yielding the correct answer of 2.67 x 10^-4 T, prompting a need for further clarification from the professor.
mbrmbrg
Messages
485
Reaction score
2

Homework Statement



In the figure (see attatched), an electron accelerated from rest through potential difference 2.0 kV enters the gap between two parallel plates having separation 20.0 mm and potential difference 130 V. The lower plate is at the lower potential. Neglect fringing and assume that the electron's velocity vector is perpendicular to the electric field vector between the plates. What uniform magnetic field allows the electron to travel in a straight line in the gap?
(answer is in Telsa, and the field is in the positive-k direction.)

Homework Equations



When a system is in equilibrium,
\sum F=0

Magnetic (Lorentz?) Force is given by:
F_B=qv\times B

Electric Force is given by:
F=qE

The electric field produced by parallel plates is:
E=\frac{V}{d}

The work-energy theorem states that W=K.
For electric potential, W=qV.
K=\frac{mv^2}{2}



The Attempt at a Solution



Finding the Velocity of the Electron as it enters the gap:
K=W
\frac{1}{2}mv^2=qV_1
where V_1 is the potential through which the electron is accelerated.
Isolate v to get
v=\sqrt{\frac{2qV_1}{m}}

Onward:
We want a situation of equilibrium where F_B=F_E.
So qv\times B=qE
the q's cancel, and the cross product is maximum, so
vB=E
plug in the velocity which we found above and the electric field of parallel plates to get
\sqrt{\frac{2qV_1}{m}}B=\frac{V_2}{d}
where V_2 is the potential between the two plates.
Isolate B to get
B = \frac{V_2}{d}\sqrt{\frac{m}{2qV_1}}

Now plug in my numbers
B = \frac{130}{20\times10^{-3}}\sqrt{\frac{9.109\times10^{-31}}{(2)(1.602\times10^{-19})(2000)}}=2.45\times10^{-4}T
Wrong:frown:

Doubly odd because when I used the values:
particle=electron
V_1=1.0kV
d=20.0mm
V_2=100V

I got the correct answer of 2.67e-4 T.
Weird...
 

Attachments

  • hrw7_28-33.gif
    hrw7_28-33.gif
    1.8 KB · Views: 1,109
Last edited:
Physics news on Phys.org
I can't find your mistake, if there even is one.
 
Thanks, nice to know it's not another one of my "redface" errors.
I guess I'll take it up with my professor, ask him to check that WebAssign is expecting a reasonable answer.
 
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