An electron accelerated by a potential difference `V=1.0 kV` moves in a uniform magnetic field at an angle `alpha = 30^@` to the vector B whose modulus is `B=29 mT.` Find the pitch of the helical trajectory of the electron.
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Time period of the electron moving in helical path `T=(2pim)/(eB)=(2xx3.14xx(9.1xx10^-31))/((1.6xx10^-19)(29xx10^-3))=1.232xx10^-9s` The K.E. acquired by the electron is given by `1/2mv^2=eV or v=sqrt((2eV)/m)` `:. v=sqrt([(2xx(1.6xx10^-19)xx10^3)/(9.1xx10^-31)])=1.875xx10^7 ms^-1` Now, if `v_(||)` be the velocity of electron parallel to megnetic field, then `v_(||)= vcos alpha =1.875xx10^7 cos 30=1.624xx10^7ms^-1` `:.` Pitch `=v_(||)xxT=(1.624xx10^7)(1.232xx10^-9)` `=2xx10^-2m=2cm`.
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