Question 220 of 276

Proton with kinetic energy of $1MeV$ 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$0.71mT$B$7.1mT$C$0.071mT$D$71mT$

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Answer: (A) $0.71mT$

$∴K.E.=1.6\times {10}^{-13}=\dfrac{1}{2}\times 1.6\times {10}^{-27}{v}^{2}$
$v=\sqrt{2}\times {10}^{7}m/s$
$∴Bqv=ma$
$B=\dfrac{1.6\times {10}^{-27}\times {10}^{12}}{1.6\times {10}^{-19}\times \sqrt{2}\times {10}^{7}}$
$=0.71\times {10}^{-3}T$
So,
$B=0.71mT$