# Radius of helical motion in a magetic field

1. Apr 4, 2014

### wai1997401

1. The problem statement, all variables and given/known data
When the charge enters the magnetic feld at an angle x , how the radius can be expressed as?

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

3. 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 :)

2. Apr 4, 2014

### TSny

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?

3. Apr 4, 2014

### lightgrav

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".

4. Apr 4, 2014

### wai1997401

but the angle x is a θ less then 90 degrees but bigger than 0 degree...the force applied is never 0