Question

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

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

Answer 1

The correct option is A $6.5\times {10}^{15}m/se{c}^2$

The force experienced by the proton due to the magnetic field provides an acceleration to it.

$F=ma=qvB$

$\Rightarrow a=\frac{qvB}{m}$

$=\frac{1.6\times {10}^{-19}\times 2\times 3.4\times {10}^7}{1.67\times {10}^{-27}}$

$=6.5\times {10}^{15}m/sec$