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Q. A proton, a deuteron and an $\alpha$-particle accelerated through the same potential difference enter a region of uniform magnetic field, moving at right angles to $B$. What is the ratio of their K.E.?

JIPMERJIPMER 2013Moving Charges and Magnetism

Solution:

K.E. gained by charged particle of charge q when accelerated under a potential difference $V$ will be $E_k = qV$
For a given $V,E_k \propto q$
For proton, deuteron and $\alpha$-particle, the ratio of charges is 1 : 1 : 2.