An electron of 45 eV energy is revolving in a circular path in a magnetic field of intensity 9 x 10^–5 Wb m^–2. Determine (i) the speed of the electron (ii) radius of the circular path.

Solution

Energy of electron, E= 45 eV

That is,

\frac{1}{2} {mv}^{2} = 45 \times 1.6 \times 10^{- 19}

v = \sqrt{\frac{2 \times 45 \times 1.6 \times 10^{- 19}}{9.1 \times 10^{- 31}}} {ms}^{- 1} = 3.98 \times 10^{6} {ms}^{- 1}

which, is the required speed of electron.

ii) Now, since its a circular path the magnetic force is equal to the centripetal force.

Therefore,

Bev = \frac{{mv}^{2}}{r} r = \frac{mv}{Be}

r = \frac{9.1 \times 10^{- 31} \times 3.98 \times 10^{6}}{9 \times 10^{- 5} \times 1.6 \times 10^{- 19}} m = 0.25 m

is the required radius of the circular path.

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