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jeyben [28]
3 years ago
15

An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.245 T. If the kinetic energy of the electr

on is 2.90 ✕ 10−19 J, find the speed of the electron and the radius of the circular path.
Physics
1 answer:
Amiraneli [1.4K]3 years ago
8 0

Answer

Given,

Magnetic field, B = 0.245 T

KE of the electron = 2.90 x 10⁻¹⁹ J

Speed of electron = ?

KE = \dfrac{mv^2}{2}

v=\sqrt{\dfrac{2(KE)}{m}}

v=\sqrt{\dfrac{2\times 2.90\times 10^{-19}}{9.11\times 10^{-31}}}

v = 7.97 x 10⁵ m/s

radius of the circular path

so,

\dfrac{mv^2}{r}=evB

r=\dfrac{mv}{eB}

r=\dfrac{9.11\times 10^{-31}\times 7.97 \times 10^5}{1.6\times 10^{-19}\times 0.245}

r = 1.85 x 10⁻⁵ m

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