JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2020
- Medium
- MCQ
Proton with kinetic energy of \(1\;MeV\) moves from south to north. It gets an acceleration of \(10^{12}\; \mathrm{m} / \mathrm{s}^{2}\) by an applied magnetic field (west to east). The value of magnetic field :.......\(mT\) (Rest mass of proton is \(1.6 \times 10^{-27} \;\mathrm{kg}\) )
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(0.71\)
Step-by-step Solution
Detailed explanation
\(\mathrm{a}=\frac{\mathrm{qvB}}{\mathrm{m}}\)
\(\mathrm{B}=\frac{\mathrm{ma}}{\mathrm{qv}}=\frac{\mathrm{ma} \sqrt{\mathrm{m}}}{\sqrt{2 \mathrm{k}}}\)
\(=\frac{\mathrm{m}^{3 / 2} \mathrm{a}}{\mathrm{e} \sqrt{2 \mathrm{k}}}=\frac{\left(1.6 \times 10^{-27}\right)^{3 / 2} \times 10^{12}}{1.6 \times 10^{-19} \sqrt{2 \times 1 \times 10^{6} \times 1.6 \times 10^{-19}}}\)
\(=0.71 \mathrm{mT}\)
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