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Electrons emitted with negligible speed from an electron gun are accelerated through a potential difference `V` along `X`-axis.These electrons emerges from a narrow hole into a unifrom magnetic field `B` directed along the axis.However, some of the electrons emerging from the hole make slightly divergent angles as shown in figure.These paraxial electrons meet for second time on the `X`-axis at a distance `sqrt((Npi^(2)mV)/(eB^(2)))`.Then find value of `N`.(Neglect interaction between electrons) |
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Answer» Correct Answer - B::C Pitch `=(2pimcostheta)/(qB) eV=1/2mv^(2)` `=(2pisqrt(2meV))/(eB) (costheta~~1) rArr "pitch" =sqrt((8pi^(2)mV)/(eB^(2)))` `:.` Distance from point of divergence `=sqrt((32pi^(2)mV)/(eB^(2)))` |
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