Question

A proton enters a magnetic field of flux density 1.5 T velocity of $20\times {10}^{2}m{s}^{-1}$ at an angle of ${30}^{\circ }$ with the field. The force on the position is $[e_p=1.6\times {10}^{-19}C]$

• A. $2.4\times {10}^{-10}N$
• B. $2.403\times {10}^{-12}N$
• C. $3\times {10}^{-5}N$
• D. $3\times {10}^{-4}N$

Answer 1

The correct option is B $2.403\times {10}^{-12}N$

The magnitude of the magnetic force on a charge particle $F=\left|q\right|vB\sin \theta$

Here, q=charge on a proton $=1.602\times {10}^{-19}C$

$\begin{array}{c}V=2\times {10}^{7}m/s\\ \theta ={30}^0\\ B=1.5T\end{array}$

Substituting all these values in the expression for $F$

We get,

$\begin{array}{c}F=1.602\times {10}^{-19}C\times 2\times {10}^{7}m/s\times 1.5T\sin {30}^0\\ =2.403\times {10}^{-12}N\end{array}$

$\therefore$ Option $B$ is correct.