
#1
Feb2612, 06:46 PM

P: 32

1. The problem statement, all variables and given/known data
A certain solenoid(50cm long with 2000 loops) carries a current of 0.70A and is in vacuum. An electron is shot at an angle of 10deg to the solenoid axis from a point on the axis. Find the speed of the electron if it is just to miss hitting the inside of the 1.6cm diameter solenoid? 2. Relevant equations B=unI r=mv/qB v=qrB/m 3. The attempt at a solution B=(12.5*10^7)(2000/0.5m)(0.7)/sin10=0.020T v=0.20T * 0.008m*1.6*10^19C=2.8*10^7m/s<<<the manual says the velocity is 1.4*10^7 don't understand why it's half of what I got. Please help. 



#2
Feb2612, 10:25 PM

Mentor
P: 11,415

The loops that the electron will perform will not occupy the whole of the solenoid cross section. Remember that the electron is starting from the solenoid axis.




#3
Feb2612, 10:46 PM

P: 32

does that make the angle double?




#4
Feb2612, 11:41 PM

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P: 11,415

Velocity of electron shot from a solenoid. 



#5
Feb2612, 11:51 PM

P: 32

Only half a circle? Because the entire circle isnt in the entire plane?




#6
Feb2612, 11:56 PM

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P: 11,415





#7
Feb2712, 12:02 AM

P: 32

The circle's diameter appears to be .08m 



#8
Feb2712, 12:15 AM

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P: 11,415





#9
Feb2712, 12:37 AM

P: 32

0.04m so that's why.
The little circle represents the electron's circular component of velocity while r helps us determine the linear component? Then we find the component perpendicular to r 



#10
Feb2712, 01:25 AM

P: 32

Ahhh the electron acquires helical motion due to being at an acute angle to the magnetic field. That's what the little circle represents.




#11
Feb2712, 08:12 AM

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P: 11,415





#12
Feb2712, 09:33 AM

P: 32

Great!
Hopefully im not undoing everything here but the circle is half as big b/c the field lines are at r and 2r(middle of solenoid cross section). respectively? I may be overthinking just wondering why and if the diameter is attributed to those velocity components. (In short what keeps me from drawing a circle the entire diameter of the solenoid? Or any other diameter?) 



#13
Feb2712, 09:50 AM

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P: 11,415





#14
Feb2712, 09:55 AM

P: 32

Okay that's the issue then I didn't know exactly where the axis fell. My textbook isn't that good it seems. So basically the electron starts from the center there and continues in it's path as discussed. For some reason I thought it shot out from the wire in the spring. Although the path is helical it doesn't go any further than that axis point as it moves.
I was scouring the net to find a diagram on the axis! :D 


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