Answer
Given,
Magnetic field, B = 0.245 T
KE of the electron = 2.90 x 10⁻¹⁹ J
Speed of electron = ?
[tex]KE = \dfrac{mv^2}{2}[/tex]
[tex]v=\sqrt{\dfrac{2(KE)}{m}}[/tex]
[tex]v=\sqrt{\dfrac{2\times 2.90\times 10^{-19}}{9.11\times 10^{-31}}}[/tex]
v = 7.97 x 10⁵ m/s
radius of the circular path
so,
[tex]\dfrac{mv^2}{r}=evB[/tex]
[tex]r=\dfrac{mv}{eB}[/tex]
[tex]r=\dfrac{9.11\times 10^{-31}\times 7.97 \times 10^5}{1.6\times 10^{-19}\times 0.245}[/tex]
r = 1.85 x 10⁻⁵ m
