Question

The frequency of the revolution of the particle in cycles per second will be

• A. $\frac{{10}^3}{\pi \sqrt{19}}$
• B. $\frac{{10}^4}{\pi \sqrt{38}}$
• C. $\frac{{10}^4}{\pi \sqrt{19}}$
• D. $\frac{{10}^4}{2\pi \sqrt{19}}$

Answer 1

The correct option is B $\frac{{10}^4}{\pi \sqrt{38}}$

Given: $\overrightarrow{B}=(20i+30j+50k)T$

$i,j,k$ are the unit vectors.

$\left(\frac{q}{m}\right)=\frac{{10}^3}{19}C/kg$

To find: frequency of the revolution,$\nu =?$

Solution: As we know that,

charged particle's frequency in magnetic field is,

$\nu =\frac{q||\overrightarrow{B}||}{2\pi m}$

$\nu =\frac{{10}^3}{19\ast 2\pi }\sqrt{3800}$

$==>?\nu =\frac{{10}^3\ast 10}{\pi \sqrt{38}}$

$==>\nu =\frac{{10}^4}{\pi \sqrt{38}}$

hence,

The correct opt : B