A) circular but not helical
B) helical but not circular
C) neither helical nor circular
D) either helical or circular
Correct Answer: D
Solution :
When the angle between magnetic field \[\text{\vec{B}}\] and velocity vector \[\text{\vec{v}}\] is \[\text{9}{{\text{0}}^{o}},\] so the force will be maximum and always perpendicular to motion, so the path will be circular. When the electron is moving at an angle to the field (other than \[{{0}^{o}},\text{ }{{90}^{o}},\] or\[{{180}^{o}}\]), then electron moves with constant velocity \[v\cos \theta \]along the field (as no force acts on a charged particle when it moves parallel to the field) and at the same time it is also moving with velocity \[v\sin \theta \]perpendicular to the field due to which it will describe a circle (in a plane perpendicular to the field) of radius \[r=\frac{m(vsin\theta )}{qB}\] So, the resultant path will be a helix with its axis parallel to the field \[\text{\vec{B}}\]as shown in figure.You need to login to perform this action.
You will be redirected in
3 sec