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4.Moving Charges and Magnetism
medium
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
$0.71$
(JEE MAIN-2020)
Solution
$\mathrm{a}=\frac{\mathrm{qvB}}{\mathrm{m}}$
$\mathrm{B}=\frac{\mathrm{ma}}{\mathrm{qv}}=\frac{\mathrm{ma} \sqrt{\mathrm{m}}}{\sqrt{2 \mathrm{k}}}$
$=\frac{\mathrm{m}^{3 / 2} \mathrm{a}}{\mathrm{e} \sqrt{2 \mathrm{k}}}=\frac{\left(1.6 \times 10^{-27}\right)^{3 / 2} \times 10^{12}}{1.6 \times 10^{-19} \sqrt{2 \times 1 \times 10^{6} \times 1.6 \times 10^{-19}}}$
$=0.71 \mathrm{mT}$
Standard 12
Physics
