A) \[\frac{2Em}{qB}\]
B) \[\frac{\sqrt{2Em}}{qB}\]
C) \[\frac{\sqrt{Em}}{2qB}\]
D) \[\frac{\sqrt{2Eq}}{mB}\]
Correct Answer: B
Solution :
Given, Kinetic energy = E Mass = m Magnetic field = B Charge = q We know that \[F=\text{ }qvB\text{ }sin\text{ }\theta \] (motion of a charged particle in a uniform magnetic field) If \[\theta ={{90}^{o}}\] Then \[F=qvB\] ?(i) We know that also (centripetal force) \[F=\frac{m{{v}^{2}}}{r}\] ?(ii) From Eqs. (i) and (ii), we get \[qvB=\frac{m{{v}^{2}}}{r},r=\frac{mv}{qB}\left[ \because \,E=\frac{1}{2}m{{v}^{2}},v=\sqrt{\frac{2E}{m}} \right]\] \[\therefore \] \[r=\frac{m\sqrt{\frac{2E}{m}}}{qB}\Rightarrow r=\frac{\sqrt{2Em}}{qB}\]
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