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

Question

A particle of charge q and mass m is moving with a velocity - v hati ( v ≠ 0) towards a large screen placed in the Y-Z plane at a distance d. If there is a magnetic field vec B = B 0 hat k , the minimum value of v for which the particle will not hit the screen is: (A) ( qdB 0/2 m ) (B) ( qdB 0/ m ) (C) (2 qdB 0/ m ) (D) ( qdB 0/3 m )

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/12323259716b772961875cb5e0c6aabe-.png" /> 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. d =( mv / qB 0) v =( qB 0 d / m )

Q. A particle of charge $q$ and mass $m$ 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 $B = B_{0} \hat{k},$ the minimum value of $v$ for which the particle will not hit the screen is:

$\frac{q d B _{0}}{2 m}$

$\frac{q d B _{0}}{m}$

$\frac{2 q d B _{0}}{m}$

$\frac{q d B _{0}}{3 m}$

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.
$d = \frac{m v}{q B _{0}}$
$v = \frac{q B _{0} d}{m}$

Q. A particle of charge q and mass m is moving with a velocity −vi^(v=0) towards a large screen placed in the Y−Z plane at a distance d. If there is a magnetic field B=B0​k^, the minimum value of v for which the particle will not hit the screen is:

2mqdB0​​

mqdB0​​

m2qdB0​​

3mqdB0​​