Solve this following


A particle of charge $\mathrm{q}$ and mass $\mathrm{m}$ is moving with a velocity $-v \hat{i}(v \neq 0)$ towards a large screen placed in the Y-Z plane at a distance $d$. If there is a magnetic field $\overrightarrow{\mathrm{B}}=\mathrm{B}_{0} \hat{\mathrm{k}}$, the minimum value of $v$ for which the particle will not hit the screen is:


  1. $\frac{\mathrm{qdB}_{0}}{2 \mathrm{~m}}$

  2. $\frac{q d B_{0}}{m}$

  3. $\frac{2 \mathrm{qdB}_{0}}{\mathrm{~m}}$

  4. $\frac{\mathrm{qdB}_{0}}{3 \mathrm{~m}}$

Correct Option: , 2


In uniform magnetic field particle moves in a circular path, if the radius of the circular path is 'd', particle will not hit the screen.


$v=\frac{q B_{0} d}{m}$

$\therefore$ correct option is (2)


Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now