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QuestionPhysicsClass 12

What is the radius of the path of an electron (mass  $$9\times 10^{-31}kg$$ and charge  $$1.6\times 10^{-19}C$$ ) moving at a speed of  $$3\times 10^7m/s$$  in a magnetic field of  $$6\times 10^{-4}T$$  perpendicular to it? What is its frequency? Calculate its energy in $$keV$$. ( $$1\mathit{eV}=1.6\times 10^{-19}\ J$$).

Radius$$=28\ cm$$
Frequency$$=18\ \mathit{MHz}$$
Energy$$=2.5\mathit{ke}V$$
4.6
4.6

Solution

Given:
$$m=9\times 10^{-31}kg$$
$$q=1.6\times 10^{-19}C$$
$$v=3\times 10^7m/s$$
$$B=6\times 10^{-4} T$$
Radius of the path of an electron:
As we know that the radius of the circular path of a particle i.e. $$r=mv/qB$$
Using above equation we find:
$$r=\frac{mv}{qB}$$
$$=\frac{9\times 10^{-31}\mathit{kg}\times 3\times 10^7m/s}{1.6\times 10^{-19}C\times 6\times 10^{-4}T}$$
$$=28\times 10^{-2}m=28\ cm$$
Frequency:
$$\nu=\frac v{2\pi r}$$
$$\nu=\frac {3\times 10^7}{2\times 2.14 \times 28\times 10^{-2}}$$
$$=18\times 10^6Hz$$
$$=18\ \mathit{MHz}$$
Energy:
$$E=\frac 1 2mv^2$$
$$=\frac 1 2\times 9\times 10^{-31}\mathit{kg}\times 9\times 10^{14}m^2/s^2$$
$$=40.5\times 10^{-17}J$$
$$=4\times 10^{-16}J=2.5\mathit{ke}V$$