Charged particles moving through a magnetic field

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Homework Help Overview

The discussion revolves around a problem involving charged particles moving through a magnetic field, specifically focusing on calculating the velocity of an electron at an angle to the magnetic field. The magnetic field strength and the force experienced by the particle are provided.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the application of the equation F=BqVSin(Theta) to find the velocity of the particle. There are attempts to verify calculations and clarify the use of sine in the context of angles. Questions arise regarding the correctness of results and the implications of using radians versus degrees.

Discussion Status

Participants are actively engaging with the problem, offering suggestions to check calculator settings and re-evaluate calculations. There is a recognition of potential errors in the original attempts, and some participants express uncertainty about the role of the sine function in the calculations. Guidance has been provided regarding significant figures and the importance of maintaining the sine term in the equation.

Contextual Notes

There is mention of specific values for force and magnetic field strength, as well as the angle of 30 degrees, which are critical to the problem setup. Participants also note the importance of significant figures in reporting the final answer.

KieranRC
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Homework Statement


A collection of charged particles move through a magnetic field at an angle to the field lines.
Calculate the velocity of the particle if it is an electron moving at 30 degrees to the magnetic field of strength 3.4mT, causing it to experience a force of 4.7x106-18N

Homework Equations



F=BqVSin(Theta)

The Attempt at a Solution


4.7x10^-18 / ((3.410^-3)(1.602x10^-19)(Sin(30))) = 8733.44m/s
 
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Your method looks fine, but your result is incorrect. Check that your calculator is set to use degrees rather than radians for angles. Or, just recall what the value of sin(30°) is and use it: it's a very common angle and it's sine and cosine really should be memorized.
 
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gneill said:
Your method looks fine, but your result is incorrect. Check that your calculator is set to use degrees rather than radians for angles. Or, just recall what the value of sin(30°) is and use it: it's a very common angle and it's sine and cosine really should be memorized.
Okay great thank you
 
KieranRC said:

Homework Statement


A collection of charged particles move through a magnetic field at an angle to the field lines.
Calculate the velocity of the particle if it is an electron moving at 30 degrees to the magnetic field of strength 3.4mT, causing it to experience a force of 4.7x106-18N

Homework Equations



F=BqVSin(Theta)

The Attempt at a Solution


4.7x10^-18 / ((3.410^-3)(1.602x10^-19)(Sin(30))) = 8733.44m/s
Your answer is off by a factor of about 2. If you ignored the sin in the denominator you'd be close but still off by about 1%.
 
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rude man said:
Your answer is off by a factor of about 2. If you ignored the sin in the denominator you'd be close but still off by about 1%.
but if i take the sin out the don't i take out the angle altogether? and then doesn't that equation without sin just assum the angle is 90?
How do i go about correcting my answer so it is spot on?
Thanks
 
|sin(30 rad)/sin(30 deg)| ~= 2. Did you check your calculator deg/rad setting as I suggested?

Type in sin(30) right now. What do you get?
 
gneill said:
|sin(30 rad)/sin(30 deg)| ~= 2. Did you check your calculator deg/rad setting as I suggested?

Type in sin(30) right now. What do you get?
Yes i did, and in the end i git 17257.84m/s (by changing to deg)
 
KieranRC said:
but if i take the sin out the don't i take out the angle altogether? and then doesn't that equation without sin just assum the angle is 90?
How do i go about correcting my answer so it is spot on?
Thanks
No, you need to leave the sin term in. I just observed that without it you'd be close, but that was just a coincidence and to give you a hint as to what the answer might be.

Your problem, simply, is your math! And BTW you stated in your original post that " ..experience a force of 4.7x106-18N .. ". What does that mean?
BTW setting your calculator to radians instead of degrees is not the problem either.
 
rude man said:
No, you need to leave the sin term in. I just observed that without it you'd be close, but that was just a coincidence and to give you a hint as to what the answer might be.

Your problem, simply, is your math! And BTW you stated in your original post that " ..experience a force of 4.7x106-18N .. ". What does that mean?
BTW setting your calculator to radians instead of degrees is not the problem either.
The forced experienced is just what the question stated.

re-entering everything in and i get 17279.4m/s does this seem more accurate?
 
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KieranRC said:
The forced experienced is just what the question stated.

re-entering everything in and i get 17279.4m/s does this seem more accurate?
That result looks good to me. Be sure to round to the appropriate number of significant figures before submitting your result.
 
  • #11
KieranRC said:
The forced experienced is just what the question stated.

re-entering everything in and i get 17279.4m/s does this seem more accurate?
Yes, surely does!
 
  • #12
@gneill, seems your suggestion to check for rads rater than sines was correct after all. :oops:
 

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