Question

A proton $\left( m _{ p }=1.67 \times 10^{-27} kg \right.$ and $\left. q _{ p }=1.6 \times 10^{-19} C \right)$ enters perpendicular to the magnetic field of intensity $2 Wb / m ^{2}$ with a velocity $3.4 \times 10^{7} m / sec$. The acceleration of the proton should be

• A. $6.5 \times 10^{15} m / sec ^{2}$
• B. $6.5 \times 10^{13} m / sec ^{2}$
• C. $6.5 \times 10^{11} m / sec ^{2}$
• D. $6.5 \times 10^{9} m / sec ^{2}$

Answer 1

Answer: (A)

As we know that
$F = ma$
$qvB \sin 90^{\circ}= ma$
$a =\frac{ qvB }{ m }$
$a=\frac{1.6 \times 10^{-19} \times 2 \times 3.4 \times 10^{7}}{1.67 \times 10^{-27}}$
$a=6.5 \times 10^{15} m / sec$