
If a charged particle enters into a uniform magnetic field neither perpendicular nor parallel to the field. What is the path of the charged particle?
A. Straight line
B. Elliptical
C. Helical
D. Parabola
Answer
475.2k+ views
Hint:To answer this question, we will have to consider the force moving in a magnetic field. Which gives us the relation between the forces, velocity and magnetic field. And then check all the conditions and these conditions will help us to find the path of the charged particle entering a uniform magnetic field neither perpendicular nor parallel to the field.
Complete step by step answer:
Assume that the charged particle enters the magnetic field at an angle $\theta $ to the direction of the magnetic field.
So, $v\cos \theta $ will be the component of velocity along the, magnetic field, therefore the force acting on the particle due to this component of velocity is $qv\cos \theta B\sin \theta = 0$
As a result, there is no force preventing this component of velocity from slowing or changing direction, causing the charged particle to proceed in the direction of the magnetic field.
Now,$v\sin \theta $ will be the component of velocity perpendicular to the magnetic field, therefore we can say that the force acting due to this component of velocity is $qv\sin \theta B\sin {90^ \circ } = qv\sin \theta B$
As a result of this component of velocity, the charged particle tends to move in a circle while also moving forward, resulting in a motion that looks like this, which is referred to as a helical motion.
Hence, the path of the charged particle is Helicaland option C is the correct option.
Note: When a charged particle follows a helical path, it may pass through an area with a non-uniform magnetic field. Assume that the particle travels from a stronger magnetic field to a weaker magnetic field and then again back to the stronger magnetic field. Before entering the zone with a greater magnetic field, the particle may bounce back. When the magnetic field is perpendicular to the velocity, it has no effect on the charged particle, whether it is positively or negatively charged.
Complete step by step answer:
Assume that the charged particle enters the magnetic field at an angle $\theta $ to the direction of the magnetic field.
So, $v\cos \theta $ will be the component of velocity along the, magnetic field, therefore the force acting on the particle due to this component of velocity is $qv\cos \theta B\sin \theta = 0$
As a result, there is no force preventing this component of velocity from slowing or changing direction, causing the charged particle to proceed in the direction of the magnetic field.
Now,$v\sin \theta $ will be the component of velocity perpendicular to the magnetic field, therefore we can say that the force acting due to this component of velocity is $qv\sin \theta B\sin {90^ \circ } = qv\sin \theta B$
As a result of this component of velocity, the charged particle tends to move in a circle while also moving forward, resulting in a motion that looks like this, which is referred to as a helical motion.
Hence, the path of the charged particle is Helicaland option C is the correct option.
Note: When a charged particle follows a helical path, it may pass through an area with a non-uniform magnetic field. Assume that the particle travels from a stronger magnetic field to a weaker magnetic field and then again back to the stronger magnetic field. Before entering the zone with a greater magnetic field, the particle may bounce back. When the magnetic field is perpendicular to the velocity, it has no effect on the charged particle, whether it is positively or negatively charged.
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