Question

A charge particle moving in magnetic field B, has the components of velocity along B as well as perpendicular to B. The path of the charge particle will be [8 April, 2023 (Shift-I)]

• A. helical path with the axis perpendicular to the direction of magnetic field B
• B. straight along the direction of magnetic field B
• C. helical path with the axis along magnetic field B
• D. circular path

Answer 1

Answer: (C)

Helical path with the axis along magnetic field B

Answer 2

The problem asks about the path of a charged particle moving in a uniform magnetic field $\vec{B}$, where its velocity $\vec{v}$ has components both along $\vec{B}$ and perpendicular to $\vec{B}$.

Let the velocity of the charged particle be $\vec{v}$. We can decompose $\vec{v}$ into two components:

1. $\vec{v}_{\parallel}$: The component of velocity parallel to the magnetic field $\vec{B}$.
2. $\vec{v}_{\perp}$: The component of velocity perpendicular to the magnetic field $\vec{B}$.

So, $\vec{v} = \vec{v}_{\parallel} + \vec{v}_{\perp}$.

The magnetic force $\vec{F}$ on a charged particle $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is given by the Lorentz force formula: $\vec{F} = q(\vec{v} \times \vec{B})$

Substitute $\vec{v} = \vec{v}_{\parallel} + \vec{v}_{\perp}$: $\vec{F} = q((\vec{v}_{\parallel} + \vec{v}_{\perp}) \times \vec{B})$ $\vec{F} = q(\vec{v}_{\parallel} \times \vec{B}) + q(\vec{v}_{\perp} \times \vec{B})$

Let's analyze the two terms:

1. Force due to $\vec{v}_{\parallel}$: Since $\vec{v}_{\parallel}$ is parallel to $\vec{B}$, the cross product $\vec{v}_{\parallel} \times \vec{B}$ is zero. Therefore, there is no magnetic force component acting on the particle due to its velocity component along the magnetic field. The particle will continue to move with a constant velocity $\vec{v}_{\parallel}$ along the direction of $\vec{B}$.

2. Force due to $\vec{v}_{\perp}$: Since $\vec{v}_{\perp}$ is perpendicular to $\vec{B}$, the magnetic force component is $\vec{F}_{\perp} = q(\vec{v}_{\perp} \times \vec{B})$. The magnitude of this force is $F_{\perp} = q v_{\perp} B$. The direction of this force is perpendicular to both $\vec{v}_{\perp}$ and $\vec{B}$. This force acts as a centripetal force, continuously deflecting the particle and causing it to move in a circular path in the plane perpendicular to $\vec{B}$. The radius of this circular path is given by $r = \frac{m v_{\perp}}{qB}$, and the angular frequency (cyclotron frequency) is $\omega = \frac{qB}{m}$.

Combined Motion: The overall motion of the charged particle is a superposition of these two independent motions:

• A uniform linear motion along the direction of the magnetic field (due to $\vec{v}_{\parallel}$).
• A uniform circular motion in a plane perpendicular to the magnetic field (due to $\vec{v}_{\perp}$).

When a circular motion is combined with a linear motion perpendicular to the plane of the circle, the resultant path is a helix. The axis of this helix will be along the direction of the uniform linear motion, which is along the direction of the magnetic field $\vec{B}$.

Therefore, the path of the charged particle will be a helical path with its axis along the magnetic field $\vec{B}$.