Question

A charged particle of unit mass and unit charge moves with velocity $\overrightarrow{v}=(8\hat{i}+6\hat{j})m/s$ in a magnetic field of $\overrightarrow{B}=2\hat{k}T$. Choose the correct alternative(s).

• 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 3.14 s

Answer 1

Answer: (A), (B), (D)

The path of the particle may be $x^2+y^2-4x-21=0$
The path of the particle may be $x^2+y^2=25$
The time period of the particle will be 3.14 s

$r=\frac{mV}{Bq}=\frac{\left(1\right)\left(10\right)}{\left(2\right)\left(1\right)}=5m$

$T=\frac{2\pi m}{Bq}=\frac{\left(2\right)\left(\pi \right)\left(1\right)}{\left(2\right)\left(1\right)}=\pi =3.14s$

Plane of circle is perpendicular to $\overrightarrow{B}$, i.e.,X-Y plane.