Respuesta :
Explanation:
It is given that,
Magnetic field, B = 0.1 T
Acceleration, [tex]a=6\times 10^{15}\ m/s^2[/tex]
Charge on electron, [tex]q=1.6\times 10^{-19}\ C[/tex]
Mass of electron, [tex]m=9.1\times 10^{-31}\ kg[/tex]
(a) The force acting on the electron when it is accelerated is, F = ma
The force acting on the electron when it is in magnetic field, [tex]F=qvB\ sin\theta[/tex]
Here, [tex]\theta=90[/tex]
So, [tex]ma=qvB[/tex]
Where
v is the velocity of the electron
B is the magnetic field
[tex]v=\dfrac{ma}{qB}[/tex]
[tex]v=\dfrac{9.1\times 10^{-31}\ kg\times 6\times 10^{15}\ m/s^2}{1.6\times 10^{-19}\ C\times 0.1\ T}[/tex]
v = 341250 m/s
or
[tex]v=3.41\times 10^5\ m/s[/tex]
So, the speed of the electron is [tex]3.41\times 10^5\ m/s[/tex]
(b) In 1 ns, the speed of the electron remains the same as the force is perpendicular to the cross product of velocity and the magnetic field.
