JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
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 \(1.6 \times 10^{-27} \;\mathrm{kg}\) )
- A \(71\)
- B \(7.1\)
- C \(0.071\)
- D \(0.71\)
Answer & Solution
Correct Answer
(D) \(0.71\)
Step-by-step Solution
Detailed explanation
\(\mathrm{a}=\frac{\mathrm{qvB}}{\mathrm{m}}\) \(\mathrm{B}=\frac{\mathrm{ma}}{\mathrm{qv}}=\frac{\mathrm{ma} \sqrt{\mathrm{m}}}{\sqrt{2 \mathrm{k}}}\)…
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