Radius of helical motion in a magetic field

[wai1997401]

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 :)

 

[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?

 

[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".

 

[wai1997401]

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

 

[HalfLostAndSearching]

Your attempt at a solution is correct. The radius of the helical motion in a magnetic field can be expressed as r=mv/qbsinx, where m is the mass of the charge, v is its velocity, q is its charge, b is the strength of the magnetic field, and x is the angle at which the charge enters the field.

It is possible that the question meant to say that the radius is inversely proportional to sinx, rather than directly proportional. In that case, you would need to rearrange the equation to r=qbsinx/mv. However, without more context, it is difficult to say for sure if there is a mistake in the question or in your solution. It is always important to carefully read and interpret the given information in a question before attempting to solve it.