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 \(\overrightarrow{ B }= B _{0} \hat{ k },\) the minimum value of \(v\) for which the particle will not hit the screen is

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Answer & Solution

Correct Answer

(B) \(\frac{q d B_{0}}{m}\)

Step-by-step Solution

Detailed explanation

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.

\(d =\frac{ mv }{ qB _{0}}\)

\(v =\frac{ q B _{0} d }{ m }\)

\(\therefore\) correct option is \((2)\)

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