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  • A particle of change 'q' and mass 'm' is moving with a velocit towards a large screen in the Y-Z plane at a distance 'd'. if ther

A particle of change 'q' and mass 'm' is moving with a velocit -vi(v\neq 0) towards a large screen in the Y-Z plane at a distance 'd'. if there is a magnetic field \overrightarrow{B}=Bo\widehat{k}, the minimum value of v for which the particle will not hit the screen is:
Option: 1 \frac{qdBo}{m}
Option: 2 \frac{qdBo}{3m}
Option: 3 \frac{qdBo}{2m}
Option: 4 \frac{2qdBo}{m}

Answers (1)

best_answer

As velocity and magnetic field is perpendicular to each other So particle will perform uniform circular motion in x-y plane

And its radius is given as

r=\frac{m v}{q B_0}

So if  the particle will not hit the screen then

d<r=\frac{m v}{q B_0}\\V>\frac{dq B_0}{m}\\ V_{min}=\frac{qd B_0}{m}

Posted by

Deependra Verma

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