Question

Proton with kinetic energy of $1 \,MeV$ moves from south to north It gets an acceleration of $10^{12} \, m/s^2$ by an applied magnetic field (west to east). The value of magnetic field : (Rest mass of proton is $1.6 \times 10^{-27}\, kg$)

• A. 7.1 mT
• B. 71 mT
• C. 0.071 mT
• D. 0.71 mT

Answer 1

Answer: (D)

0.71 mT

Answer 2

$a=\frac{qvB}{m}$
$B=\frac{ma}{qv}=\frac{ma\sqrt{m}}{\sqrt{2k}}$
$=\frac{m^{3/2}a}{e\sqrt{2k}}=\frac{\left(1.6\times10^{-27}\right)^{3/2}\times10^{12}}{1.6\times10^{-19}\sqrt{2\times1\times10^{6}\times1.6\times10^{-19}}}$
$=0.71\,m\,T$