A positively charged particle (q) travelling at 30° with respect to the direction of magnetic field of strength 2.4\times {{10}^{-6}}T experiences a force of 4.8\times {{10}^{-19}} N. The speed of charged particle will be
A positively charged particle (q) travelling at 30° with respect to the direction of magnetic field of strength 2.4\times {{10}^{-6}}T experiences a force of 4.8\times {{10}^{-19}} N. The speed of charged particle will be

Solution: Option 2 is correct

We\,know\,that:

F=Bqvsin\theta

\Rightarrow v=\frac{F}{Bqsin\theta }

Putting\,the\,given\,values,\,we\,get:

v=\frac{4.8\times {{10}^{-19}}}{2.4\times {{10}^{-6}}\times 1.6\times {{10}^{-19}}\times \frac{1}{2}}

v=2.5\times {{10}^{6}}\,m/s