Question

An electron is accelerated through a PD of 100 V and then enters a region where it is moving perpendiculasr to a magnetic fiedl B=0.2T. Find the radius of the circular path. Repeat this problem for a proton.

Answer 1

Explanation:

We Have :-

Electron is accelerated through a PD of 100 V

Moving perpendicular to a magnetic field B=0.2T

To Find :-

Radius of the circular path

Solution :-

We Have

$kinetic \: energy \: of \: electron = 100e \\ \\ \\ kinetic \: energy = \dfrac{1}{2} mv^{2} \\ \\ \\ 100e =\dfrac{1}{2} mv^{2} \\ \\ \\ v = \sqrt{ \dfrac{200e}{m} } \\ \\ \\ this \: provides \: centripetal \\ \\ \\ \dfrac{mv^{2} }{r} = evb \\ \\ \\ mv = ebr \\ \\ \\ r = \dfrac{mv}{eb} \\ \\ \\ putting \: the \: values \: we \: get \\ \\ \\ r = \frac{9.1 \times 10^{ - 31} \times \sqrt{ \dfrac{200e}{m} } }{1.6 \times 10^{ - 19} \times 0.2} \\ \\ \\ solving \: this \: we \: get \\ \\ \\ r_{e} = 0.0167cm \\ \\ \\ and \: for \: proton \: putting \: the \: values \: of \: proton \: in \: the \: formula \: we \: get \\ \\ \\ r_{p} = 0.7cm$