Question

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

• 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

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

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

$r=\frac{mv}{qB}$
$=\frac{\sqrt{8^2+6^2}}{2}$
$=5$
Both $x^2+y^2$ and $x^2+y^2-4x-21=0$ are possible because Magnetic field is towards the positive z axis .
$T=\frac{2\pi m}{qB}$
$=\frac{2\pi }{2}$
$=\pi$
Particle will move in x-y plane with radius equals to $5$