WBJEE · Physics · Magnetic Effects of Current
A proton of mass \(m\) and charge \(q\) is moving in a plane with kinetic energy \(E\). If there exists a uniform magnetic field \(B\), perpendicular to the plane of the motion, the proton will move in a circular path of radius
- A \(\frac{2 E m}{q B}\)
- B \(\frac{\sqrt{2 E m}}{q B}\)
- C \(\frac{\sqrt{E m}}{2 q B}\)
- D \(\sqrt{\frac{2 E q}{m B}}\)
Answer & Solution
Correct Answer
(B) \(\frac{\sqrt{2 E m}}{q B}\)
Step-by-step Solution
Detailed explanation
Given, Kinetic energy \(=E\) Mass \(=m\) Magnetic field \(=B\) Charge \(=q\) We know that \[ F=q v B \sin \theta \] (motion of a charged particle in a uniform magnetic field) If \[ \theta=90^{\circ} \] Then \[ F=q v B \] We know that also (centripetal force)…
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