Answer:
Acceleration, [tex]a=9.36\times 10^{18}\ m/s^2[/tex]
Explanation:
It is given that,
Speed of electron, [tex]v=8\times 10^6\ m/s[/tex]
Charge on an electron, [tex]q=1.6\times 10^{-19}\ C[/tex]
Mass of electron, [tex]m=9.1\times 10^{-31}\ kg[/tex]
Magnetic field, [tex]B=5.5i-3.7j[/tex]
Magnitude, [tex]|B|=\sqrt{5.5^2+(-3.77)^2}=6.66\ T[/tex]
Magnetic force is given by :
[tex]F=qvB[/tex]
Also, F = ma
[tex]a=\dfrac{qvB}{m}[/tex]
[tex]a=\dfrac{1.6\times 10^{-19}\times 8\times 10^6\times 6.66}{9.1\times 10^{-31}}[/tex]
[tex]a=9.36\times 10^{18}\ m/s^2[/tex]
So, the acceleration of the electron is [tex]9.36\times 10^{18}\ m/s^2[/tex]. Hence, this is the required solution.
