Answer:
The magnitude of the acceleration of the electron at this point is [tex]3.94\times10^{14}\ m/s^2[/tex].
Explanation:
Given that,
Velocity [tex]v= 0.949\times10^{6}\ m/s[/tex]
Angle = 61.5°
Magnetic field = 0.01 T
We need to calculate the magnetic force
Using formula of magnetic force
[tex]F=Bev\cos\theta[/tex]
Where, B = magnetic field
v = velocity
e = charge of electron
Put the value into the formula
[tex]F=0.01\times1.6\times10^{-19}\times0.949\times10^{6}\cos61.5[/tex]
[tex]F=3.59\times10^{-16}\ N[/tex]
We need to calculate the acceleration
Using newton's second law
[tex]F= ma[/tex]
[tex]a = \dfrac{F}{m}[/tex]
Put the value into the formula
[tex]a=\dfrac{3.59\times10^{-16}}{9.1\times10^{-31}}[/tex]
[tex]a=3.94\times10^{14}\ m/s^2[/tex]
Hence, The magnitude of the acceleration of the electron at this point is [tex]3.94\times10^{14}\ m/s^2[/tex].
