The radius of the path of an electron moving at a speed of $3.2 imes 10^7 \mathrm{~ms}^{-1}$ in a magnetic field of $6 imes 10^{-4} \mathrm{~T}$ perpendicular to it is (mass of electron is $9 imes 10^{-31} \mathrm{~kg}$ and charge of electron is $\left.1.6 imes 10^{-19} \mathrm{C} ight)$

Question

The radius of the path of an electron moving at a speed of $3.2 	imes 10^7 \mathrm{~ms}^{-1}$ in a magnetic field of $6 	imes 10^{-4} \mathrm{~T}$ perpendicular to it is (mass of electron is $9 	imes 10^{-31} \mathrm{~kg}$ and charge of electron is $\left.1.6 	imes 10^{-19} \mathrm{C}ight)$

Aicharya · Accepted Answer

$\mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{9.1 	imes 10^{-31} 	imes 3.2 	imes 10^7}{1.6 	imes 10^{-19} 	imes 6 	imes 10^{-4}}=30 \mathrm{~cm}$

The radius of the path of an electron moving at a speed of $3.2 \times 10^7 \mathrm{~ms}^{-1}$ in a magnetic…

Solution