An electron is accelerated from rest by a potential difference of 435 V. It then enters a uniform magnetic field of magnitude 167 mT with its velocity perpendicular to the field. Calculate (a) the speed of the electron and (b) the radius of its path in the magnetic field.

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onyebuchinnaji onyebuchinnaji · Answer 1 · 2020-10-24T03:46:10+02:00

Answer:

(a) the speed of the electron is 1.237 x 10⁷ m/s

(b) the radius of its path in the magnetic field is 4.21 x 10⁻⁴ m

Explanation:

Given;

potential difference, V = 435 V

magnetic field, B = 167 mT = 0.167 T

(a) the speed of the electron

eV = ¹/₂mv²

[tex]v = \sqrt{\frac{2eV}{m} }\\\\ v = \sqrt{\frac{2(1.6*10^{-19})(435)}{(9.1*10^{-31})}}\\\\v = 1.237*10^7 \ m/s[/tex]

(b) the radius of its path in the magnetic field

[tex]r = \frac{mv}{eB}\\\\r = \frac{(9.1*10^{-31})(1.237*10^7)}{(1.6*10^{-19})(0.167)}\\\\r = 4.21 *10^{-4} \ m[/tex]