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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

  1. A \(\frac{ q d B _{0}}{2 m }\)
  2. B \(\frac{q d B_{0}}{m}\)
  3. C \(\frac{2 q d B_{0}}{m}\)
  4. D \(\frac{q d B_{0}}{3 m}\)
Verified Solution

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 \((2)\)