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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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