Max distance of rectangle that electron beam can pass through

AI Thread Summary
The discussion centers on determining the maximum width (d) of a rectangular region that an electron beam can pass through while subjected to a uniform magnetic field. The electrons, moving at a speed of 6.00 × 10^6 m/s in the +x direction, will follow a circular path due to the magnetic field directed in the -z direction. The key point is that if the radius of the circular path (r) is smaller than the width (d), the electrons will not exit the region and will instead turn back. The relevant equations involve the magnetic force and centripetal acceleration, specifically F=qvB and a=v^2/r. To solve for d, one must relate the radius of the electron's path to the width of the rectangular region.
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Homework Statement



A beam of electrons is fired into a rectangular region of space that contains a uniform magnetic field in the -z direction. The electrons are moving in the +x direction, as shown. The speed of the electrons in the beam is 6.00 × 106 m/s. The mass of an electron is me = 9.11 × 10−31 kg. The magnitude of the magnetic field in the rectangular region of space is 1.50 × 10−2 T. The rectangular region has a width d.


What is the maximum value of d for which the electron beam will make it through to the other side of this rectangular region, and continue on to the right of the region?


Homework Equations



|⃗v|=|E⃗|/|B⃗1|

(Equations provided)
|F⃗m| = |q||⃗v||B⃗ || sin(θ)|
|F⃗m| = IL|B⃗ || sin θ| r = m | ⃗v |
|q||B⃗ | r = m | ⃗v ⊥ | |q||B⃗ |
|F⃗E|=|F⃗B| −→ q|E⃗|=q|⃗v||B⃗1| −→ |⃗v|=|E⃗|/|B⃗1| | m⃗ | = N I A
τ = NIA|B⃗||sinθ| = |m⃗ ||B⃗||sinθ|
| B⃗ | = μ 0 I 2πr
| B⃗ | = μ 0 N I 2r
| B⃗ | = μ 0 N I L
Φv =|⃗v|Acos(θ)
ΦB =|B⃗|Acos(θ)
E = N|∆Φ| ∆t
I = V/R
I = E/R (linear DC generator)
E = BvL (linear DC generator)
I = E/R = BvL/R (linear DC generator)
Fm = ILB = (B2vL2)/R (linear DC generator)
Pmechanical = Fappl v Pelectrical = IV = IE
ΦB = BA cos(ωt) (rotary AC generator)
E = (NBAω)sin(ωt) (rotary AC generator)
I = E/R = (NBAω/R) sin(ωt) 1Tesla(1T)=1 N
μ0 =4π×10−7 Tm A



The Attempt at a Solution



E=6.00 x 10^6m/s x 1.50 x 10^-2 T
90000 T x m/s

Ok the problem I am having is relating distance to any equation I have. In previous problems I have worked, none of them had distance as part of the problem. If someone could just give me a hint or something about how to relate distance to the problem, that would be much appreciated.
 

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Magnetic fields make moving charged particles move in circles. If the radius of the circle is too small, the electrons will remain in the region.
The relevant equations are F=qvB and a=v2/r, and of course, F=ma
 
I attached a picture to go along with my problem, did you happen to look at it? I am not sure if I attached it right.
 
The picture clarified it a bit, but nothing has changed.
The electrons will follow a circle with radius r. If r is smaller than d, then the electrons will turn around and come back the way they came.
 
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