Question

An electric having 450v of energy move at right angle to a unifoy magnetic field fluid density 1.50x10-3 show that the path of the electron is circle and find its rad

Answer 1

As per the given question,

Applied potential (V)=450V

Charge on electron $(e)=1.6\times 10^{-6}C$

$B=1.5 \times 10^{-3}T$

We know that,

$qv\times B=\frac{mv^2}{R}$

$\Rightarrow qv\times B=\frac{mv^2}{R}$

$\Rightarrow R=\frac{1}{qB}\sqrt{\frac{2m^2 v^2}{2}}$

$\Rightarrow qv\times B=\frac{2eV}{R}$

$\Rightarrow R=\frac{\sqrt{2 m_e V}}{\sqrt{e}\times B}$

$\Rightarrow R = \frac{\sqrt{2\times 9.1 \times 10^{-31}\times 450}}{\sqrt{1.6\times 10^{-19}}\times 1.5\times 10^{-3}}=\frac{\sqrt{8190\times 10^{-31}}}{1.89\times 10^{-3}}$

$\Rightarrow R=\frac{28.61\times 10^{-15}}{1.89\times 10^{-3}}m$

$\Rightarrow R = 15.13\times 10^{-12} m$