Question

An electron having charge $1.6\times {10}^{-19}\text{C}$ and mass $9\times {10}^{-31}\text{kg}$ is moving in a circular orbit with a speed of $4\times {10}^6{\text{ms}}^{-1}$ in a magnetic field of $2\times {10}^{-1}\text{T}.$ The force acting on electron and the radius of the circular orbit will be:-

• A. $12.8\times {10}^{-13}\text{N},1.1\times {10}^{-4}\text{m}$
• B. $1.28\times {10}^{-14}\text{N},1.1\times {10}^{-3}\text{m}$
• C. $1.28\times {10}^{-15}\text{N},1.1\times {10}^{-5}\text{m}$
• D. $1.28\times {10}^{-12}\text{N},1.1\times {10}^{-6}\text{m}$

Answer 1

The correct option is A $12.8\times {10}^{-13}\text{N},1.1\times {10}^{-4}\text{m}$

As, the electron moves in a circular path, thus $\overrightarrow{v}\perp \overrightarrow{B}$

Magnetic force on the charge particle is,

$F_m=qvB\sin \theta =1.6\times {10}^{-19}\times 4\times {10}^6\times 2\times {10}^{-1}$

$=1.28\times {10}^{-13}\text{N}$

Radius of the circular path traced by the charge particle in the magnetic filed is,

$r=\frac{mv}{qB}=\frac{9\times {10}^{-31}\times 4\times {10}^6}{1.6\times {10}^{-19}\times 2\times {10}^{-1}}=1.1\times {10}^{-4}\text{m}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.