Magnetic force relation with v and r?

[Hardik Batra]

(1) Magnetic force acting on charge moving in magnetic field is given by,
F = qvBsinθ...F ∝ v
This means that if moving charge particle has more velocity then more magnetic force will act on it.
Due to more magnetic force the direction of velocity changes rapidly and it will move on the circular path of smaller radius.

(2). Now, Radius of circular path , R = mv/qB.
Here, R ∝ v
This equation implies that if velocity of charge particle is more then it will move on a circular path of larger radius.

In the first situation if v is more then it will move on a circular path of smaller radius.
and in the second situation if v is more then it will move on a circular path of larger radius.

There is a contradiction or am i misleading?
Please correct me if i wrong.

 

[Doc Al]

This means that if moving charge particle has more velocity then more magnetic force will act on it.

True.

Due to more magnetic force the direction of velocity changes rapidly and it will move on the circular path of smaller radius.

False. Realize that since it's moving faster, even more force is needed to keep it moving in a circle of a given radius.

 

[Hardik Batra]

Thanks.

 

[Micromike]

There are two different situations. In the first case we are visualising the relationship between the magnetic force acting on the body due to the velocity. in other words, the force depends on the velocity. The second relation keeps F constant and describes the motion of the charge in the B field. Hope that helps you

 

[olgerm]

$\begin{cases} F=q \cdot v \cdot B \cdot sin(θ)\\ a=\frac{F}{m}\\ a=\frac{v^2}{r}\\ \end{cases}\Rightarrow r=v \cdot \frac{m}{q \cdot B \cdot sin(θ)}$

θ(angle between B-field and v (speed)) has to be 90° aka $\frac{2\pi}{4}$, because otherwise F(force), θ nor r(radius), were not constant.