JEE Main · PhysicsHard
A particle of mass $m$ and charge $q$ is projected with velocity $v$ into a region of uniform magnetic field $B$ directed perpendicular to the velocity. If the particle undergoes a collision and its velocity becomes $v/2$, what is the new radius $R'$ of its circular path?
- A.$\frac{mv}{qB}$
- B.$\frac{mv}{2qB}$
- C.$\frac{mv}{4qB}$
- D.$\frac{2mv}{qB}$
Show correct answer & step-by-step solution
Correct answer: B — $\frac{mv}{2qB}$
Solution
The radius is . With , the new radius is .
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