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an electron is accelerated through a potential difference of 5kv and enters a uniform magnetic field of 0.02 wb/m2 acting normal to the direction of electron motion. Determine radius of the path. - Applied Physics 2

Short Note

an electron is accelerated through a potential difference of 5kv and enters a uniform magnetic field of 0.02 wb/m2 acting normal to the direction of electron motion. Determine radius of the path. 

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Solution

for the case of acceleration due to electric field
                                              `1 / 2 mv^2 = eV`

                                              `V = sqrt((2e)/m xx V`

                              And for the case of transverse field

                                             ` BeV = (mV^2 )/ R`
                                             `R = (m/e)(V/B)`

                                            `= (m/e)(V/B)sqrt((2e)/m xx v)`

                                          `=1/B sqrt((2xx9.1xx10^(-31) xx 5xx10^3)/(1.6 xx 10^(-19)`

                                           = 0.012 m
                                        R = 12mm

 

Concept: Motion of Electron in Electric Field
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