In Thomson's experiment to determine e/m of an electron, it is found that an electron beam having kinetic energy of 45.5 eV remains undeflected, when subjected to crossed electric and magnetic fields. If `E=1xx10^(3)Vm^(-1)`, the value of B is (mass of the electron is `9.1xx10^(-31)kg`)

Options

A. `2.5xx10^(-3) Wb m^(-2)`
B. `5.0xx10^(-4) Wb m^(-2)`
C. `2.5xx10^(-4) Wb m^(-2)`
D. `1.0xx10^(-4) Wb m^(-2)`

Answer

C

Explanation

`E=BV`
`impliesB=E/V`
`K=(1)/(2)my^(2)`
`V^(2)=(2xx45.5xx1.6xx10^(-19))/(9.1xx10^(-31))=16xx10^(12)`
`v=4xx10^(6)`
`B=(1xx10^(3))/(4xx10^(6))=2.5xx10^(-4)Wb m^(-2)`