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  • A proton with kinetic energy of  moves from south to north. It gets acceleration of 

A proton with kinetic energy of 1MeV moves from south to north. It gets acceleration of 10^{12}m/s^2 by an applied magnetic field (west to east). The value of magnetic field ( in mT):  (Rest mass of a proton is 1.6\times 10^{-27}kg)
Option: 1 0.71
Option: 2 71
Option: 3 0.071
Option: 4 7.1
 

Answers (1)

best_answer

 

   \begin{array}{l}{\because \mathrm{K.E.}=1.6 \times 10^{-13}=\frac{1}{2} \times 1.6 \times 10^{-27} \mathrm{V}^{2}} \\ \\ {\mathrm{V}=\sqrt{2} \times 10^{7}} \\ \\ {\therefore \mathrm{Bqv}=\mathrm{ma}} \\ \\ {\mathrm{B}=\frac{1.6 \times 10^{-27} \times 10^{12}}{1.6 \times 10^{-19} \times \sqrt{2} \times 10^{7}}} \\ \\ {=0.71 \times 10^{-3} \mathrm{T}} \\ \\ {\text { so } 0.71 \mathrm{mT}}\end{array} 

So option (1) is correct

Posted by

vishal kumar

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