JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
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
- A \(\frac{ q d B _{0}}{2 m }\)
- B \(\frac{q d B_{0}}{m}\)
- C \(\frac{2 q d B_{0}}{m}\)
- D \(\frac{q d B_{0}}{3 m}\)
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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