Question

A charged particle of unit mass and unit charge moves with velocity $\overrightarrow {\text{v}} = \left( {8\widehat i + 6\widehat j} \right)m{s^{ - 1}}$ in a magnetic field of $\overrightarrow B = 2\widehat kT$. 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:
${F_C} = \dfrac{{m{{\text{v}}^2}}}{r}$
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
$\overrightarrow {\text{v}} = \left( {8\widehat i + 6\widehat j} \right)m{s^{ - 1}}{\text{ }}...{\text{(i)}}$
The magnetic field is given as
$\overrightarrow B = 2\widehat kT{\text{ }}...{\text{(ii)}}$
This means that force acting on particle is
$F = qVB\sin 90 = qVB{\text{ }}...{\text{(iii)}}$
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
$
  \dfrac{{m{V^2}}}{r} = qVB \\
   \Rightarrow r = \dfrac{{mV}}{{qB}} \\
$
Inserting the known values, we get the radius to be
$r = \dfrac{{mV}}{{qB}} = \dfrac{{1 \times \sqrt {{{\left( 8 \right)}^2} + {{\left( 6 \right)}^2}} }}{{1 \times 2}} = 5$
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
$T = \dfrac{{2\pi r}}{V} = \dfrac{{2\pi \times 5}}{{10}} = \pi {\text{ s = 3}}{\text{.14s}}$
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 $\theta = {90^0}$.