Question

A charged particle of unit mass and unit charge moves with velocity $\overrightarrow{\text{v}}=\left(8\hat{i}+6\hat{j}\right)m{s}^{-1}$ in a magnetic field of $\overrightarrow{B}=2\hat{k}T$. Choose the correct alternatives(s).
This question has multiple correct options
(A). The path of the particle may be $x^2+y^2-4x-21=0$
(B). The path of the particle may be $x^2+y^2=25$
(C). The path of the particle may be $y^2+z^2=25$
(D). The time period of the particle will be $3.14s$

Answer 1

Hint: A charged particle moves in a circular trajectory in the region containing a magnetic field. The necessary centripetal force required to move in circular motion is provided by the magnetic force acting on the charged particle.

Formula used:
Formula for centripetal force is:
$F_C={\displaystyle \frac{m{\text{v}}^2}{r}}$
where $F_C$ is the centripetal force which makes the particle of mass m move in a circular orbit of radius r with velocity v.
Magnetic force acting on a charged particle:
$F=q\overrightarrow{V}\times \overrightarrow{B}=qVB\sin \theta$

Complete step-by-step answer:
We are given the velocity of the charge particle to be
$\overrightarrow{\text{v}}=\left(8\hat{i}+6\hat{j}\right)m{s}^{-1}...\text{(i)}$
The magnetic field is given as
$\overrightarrow{B}=2\hat{k}T...\text{(ii)}$
This means that force acting on particle is
$F=qVB\sin 90=qVB...\text{(iii)}$
Also the charged particle moves in a circular orbit when velocity is perpendicular to the magnetic field.
The necessary centripetal force is provided by the magnetic force acting on the particle, therefore we can write that
$$\begin{gathered} {\displaystyle \frac{m{V}^2}{r}}=qVB \\ \Rightarrow r={\displaystyle \frac{mV}{qB}} \end{gathered}$$
Inserting the known values, we get the radius to be
$r={\displaystyle \frac{mV}{qB}}={\displaystyle \frac{1\times \sqrt{{\left(8\right)}^2+{\left(6\right)}^2}}{1\times 2}}=5$
The option A says that path of the particle is $x^2+y^2-4x-21=0$
This equation can be re-written as ${\left(x-2\right)}^2+y^2=5^2$ which is the equation of circle of radius 5 same as the obtained value above. So, option A is correct.
The option B is also correct as the radius of the given circle is 5.
The option C is wrong because the particle will move in the x-y plane perpendicular to the direction of the magnetic field.
The time period of particle is calculated as
$T={\displaystyle \frac{2\pi r}{V}}={\displaystyle \frac{2\pi \times 5}{10}}=\pi \text{ s = 3}\text{.14s}$
Hence, option D is also correct.

Note: From the unit vectors involved in the expression for velocity and magnetic field, we can judge that the particle is moving in x-y plane whereas the magnetic field is in z-direction. This implies that the velocity and magnetic field are perpendicular to each other and $\theta ={90}^0$.