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JEE Mains · Physics · STD 12 - Moving charges and magnetism
  • JEE Mains 2020
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  • 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 \(1.6 \times 10^{-27} \;\mathrm{kg}\) )

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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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