mbrmbrg
Apr16-07, 08:27 PM
1. The problem statement, all variables and given/known data
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.)
2. Relevant 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}
3. 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...
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.)
2. Relevant 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}
3. 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...