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An electron beam with electron speed of ...

An electron beam with electron speed of `5.2 xx 10^(4) ms^(-1)` is subjected to a magnetic field of `1.3 xx 10^(-6)T` normal to the beam. If the radius of path traced by beam is `22.7 xx 10^(-2)m`. Calculate the electronic charge. Gives, mass of electron `=9.1 xx 10^(-31)kg`.

Text Solution

Verified by Experts

`evB= (mv^(2))/(r ) therefore e= (mv)/(Br)`
`m= 9.1 xx 10^(-31)kg`
`v= 5.2 xx 10^(4) ms^(-1)`
`B= 1.3 xx 10^(-6)T`
`r= 22.7 xx 10^(-2)m`
`e= (9.1 xx 10^(-31) xx 5.2 xx 10^(4))/(1.3 xx 10^(-6) xx 22.7 xx 10^(-2))I`
`=1.6035 xx 10^(-19)C`
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