# What is the radius of the helix a charged particle makes when entering a magnetic field at an angle?

Here is the equation I found in my textbook but it doesn't make sense: $$R=\frac{mv\sin α}{QB}$$ . Looking at this formula we can say that particle A moving through magnetic field at an angle $$α<β$$ would follow a helical path with radius $$r_1 and particle B moving through the same field with an angle $$β$$ would follow a helical path with radius $$r_2$$. But we know that force on a particle with smaller angle is smaller than on a particle with a bigger angle and this formula hence doesn't make sense?

• Thank your for the correction. English is not my native language so I didn't notice the difference. – ToTheSpace 2 May 23 at 16:01

The centripetal acceleration required to hold an object in a circular path is $$a_c=\frac{v^2}{r}$$. What matters here is the projection of the spiral path (which is a circle) and the component of $$v$$ perpendicular to $$B$$, which is $$v_\perp=v\sin\theta$$. So required acceleration for circular path is $$a_c=\frac{v^2 \sin^2\theta}{r}$$. Put it $$r$$: $$a_c=\frac{v^2 \sin^2\theta q B}{m v \sin\theta} = \frac{v \sin\theta q B}{m}$$, which is consistent. Yes, force and acceleration decrease with angle, but it's because the particle isn't going as fast around the circle part of the spiral. It has more velocity along the linear component of the spiral path, which takes no force to maintain.