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## Q. 3.10

When an electron is placed in a magnetic field with a period of rotation $T=\frac{35.5}{B}\times 10^{-12}$, so that the trajectory of an electron is a circle.

(i) What is the radius described by an electron placed in a magnetic field, perpendicular to its motion, when the accelerating potential is 900 V, and $B = 0.01 Wb/m^{2}$? (ii) What is the time period of rotation?

## Verified Solution

Given                        $T=\frac{35.5}{B}\times 10^{-12} s,B=Wb/m^{2},V_{a}=900V$

Therefore,                $T = 3.55 × 10^{–9} s$

Velocity,                    $v=\sqrt{\frac{2qV_{a}}{m} }=\sqrt{2\times 1.76\times 10^{11}\times 900 }=17.799\times 10^{6} m/s$

Radius,                       $r=\frac{mv}{qB}=\frac{v}{(qlm)B}=\frac{17.799\times 10^{6} }{1.76\times 10^{11}\times 0.01 }=10.11mm$