Question

A proton is fired with a speed of $2\times {10}^6$ m/s at an angle of ${60}^o$ to the X- axis. If a uniform magnetic field of $0.1T$ is applied along the Y- axis, the force acting on the proton is

• A. $1.63\times {10}^{-14}N$
• B. $1.6\times {10}^{-14}N$
• C. $3.23\times {10}^{-14}N$
• D. $3.2\times {10}^{-14}N$

Answer 1

The correct option is B $1.6\times {10}^{-14}N$

$F=q(\overrightarrow{V}\times \overrightarrow{B})$
$\theta ={30}^o$ angle between velocity & field
$=1\cdot 6\times {10}^{-19}\times 2\times {10}^6\times 0\cdot 1\times \sin {30}^o$
$=1\cdot 6\times {10}^{-14}N$