Q114 / 180 · AIIMS 2011 · Physics
A proton travels few distance in an electric field, then it enters a crossed magnetic field of $1 \mathrm{~T}$ and radius $0.2 \mathrm{~m}$, find the velocity of proton.
- Correct answerA$0.2 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}$
- B$0.2 \times 10^7 \mathrm{~m} \mathrm{~s}^{-1}$
- C$0.2 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}$
- D$2 \times 10^7 \mathrm{~m} \mathrm{~s}^{-1}$
Solution
Given, $B=1 \mathrm{~T}, r=0.2 \mathrm{~m}$
$$
r=\frac{m v}{q B}
$$
i.e, $\quad v=\frac{0.2 \times 1.6 \times 10^{-19} \times 1}{1.6 \times 10^{-27}}=0.2 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}$
Hence, the correct answer is option: 'A'.