Q 12-04-001JEE AdvancedTop questionHard
A proton moves at $2 \times 10^6\ \text{m/s}$ perpendicular to a uniform magnetic field of $0.1\ \text{T}$. Find the radius of its circular path in cm. (Take $m_p = 1.67\times10^{-27}\ \text{kg}$, $e = 1.6\times10^{-19}\ \text{C}$.)
Numerical value type. Enter your answer.
Answer: 20.9
The magnetic force provides the centripetal force: $qvB = \dfrac{mv^2}{r}$.
$$r = \frac{mv}{qB} = \frac{(1.67\times10^{-27})(2\times10^{6})}{(1.6\times10^{-19})(0.1)} \approx 0.209\ \text{m} = 20.9\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics