Question

A neutron, a proton, an electron and alpha particle enter a region of the constant magnetic field with equal velocities. The magnetic field is along the inward normal to the plane of the paper, the tracks of the particle are labelled in figure, the electron follows track __ and alpha particle follows track____
A. A, C
B. C, A
C. D, B
D. B, D

Answer 1

We will use the concept of magnetic force formula which determines the trajectory of a charged particle moving inside the magnetic field. And also neutron particles have no charge whereas protons and electrons have positive and negative charge whereas alpha particles have positive charge.

Complete step by step answer:
Firstly we need to understand the magnetic force formula which is given as,
$\overrightarrow F = q(\overrightarrow v \times \overrightarrow {B)}$
where $F$ is the force, $v$ is velocity and $B$ is magnetic field.
We have given that $\overrightarrow v$ and $\overrightarrow B$ are perpendicular to each other since magnetic field is in downward direction so term $(\overrightarrow v \times \overrightarrow {B)}$ will be simply as $vB\sin {90^ \circ }$ which results
$F = qvB$. And Force will be in direction right to the initial point if charge is negative and left to the initial point if charge is positive. Using Fleming Right hand rule.

Now, Let us check for electron, since electron has a negative charge so force will act in right to the initial point so best path is described by D. $F = - qvB$. For alpha particles, since alpha particle and proton both have positive charge so their path must be in left but as we know greater the charge greater the deflection and we know that alpha particle has greater charge than proton so path must be followed by B. $F = + 2qvB$.

Hence, the correct option is C.

Note: Fleming right hand rule states “point index finger in the direction of magnetic field and thumb in the direction of velocity of particle then middle finger gives us the direction of force”. Here, the neutron has no charge so force on the neutron will be zero so its path is shown by C (no deflection).