Radius of helical motion in a magetic field

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Homework Help Overview

The discussion revolves around the radius of helical motion of a charged particle in a magnetic field, particularly focusing on how the radius relates to the angle at which the charge enters the field.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive the radius using the relationship between magnetic force and centripetal force, leading to a formula for radius. Some participants question the implications of the angle being zero and whether the derived formula aligns with the proportionality stated in the problem.

Discussion Status

The discussion is ongoing, with participants exploring the implications of the angle on the radius and questioning the assumptions made in the original poster's reasoning. There is no explicit consensus, but various interpretations and considerations are being examined.

Contextual Notes

Participants note that the angle x is constrained to be greater than 0 degrees and less than 90 degrees, which affects the force applied to the moving charge.

wai1997401
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Homework Statement


When the charge enters the magnetic feld at an angle x , how the radius can be expressed as?


Homework Equations


Force of charge in a magentic field = qvbsinx
centripetal force = mv^2/r

The Attempt at a Solution


when i combined the two equations above, i got r=mv/qbsinx,
but the question i did said the radius is directly proportional to sinx. What i have done wrong?Hope someone can help me :)
 
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Seems to me your answer is correct. What if the angle x = 0? What does your answer give for r for this case? Does it make sense?

What would r be for x = 0 if r were proportional to sinx? Would that make sense?
 
comment: 2) Force applied to the moving charge is q v B sinθ ... Force "of" the charge sounds like the charge is applying that Force.

If the Force is zero, the curvature is zero ... curvature is proportional to sinθ.
I hope you mis-read the question's "discussion".
 
but the angle x is a θ less then 90 degrees but bigger than 0 degree...the force applied is never 0
 

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