Question

an electron accelerated from rest through a voltage of 770 v enters a region of constant magnetic field. part a if the electron follows a circular path with a radius of 18 cm , what is the magnitude of the magnetic field?

Answer 1

The magnitude of the magnetic field on which an electron accelerated from rest through a voltage of 770v enters a region of constant magnetic field is 3.744 * $10^{-4}$ T

From the conservation of energy, we have

$\frac{1}{2}mv^{2} =qV$

$v=\sqrt{\frac{2qV}{m} }$

$=\sqrt{\frac{2*(1,6 * 10^{-19})(770) }{9.11 * 10^{-31} } }$

=$1.644*10^{7$m/s

At Equilibrium, Centripetal force = magnetic force

$\frac{mv^{2} }{r}=Bqv$

B=mv/rq

by putting, m=9.11* 10^-31 kg

v=1.664*10^7m/s

r=25*10^-2m

q=1.6*10^-19 C

We get, B=3.774* 10^-4 T

Hence the magnitude of the magnetic field, B=3.774* 10^-4 T

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