Answer:
a) 1.38°
b) 7.53*10^11 m/s/s
c) 6.52*10^-9m
Explanation:
a) to find the angle you can use the dot product between two vectors:
[tex]\vec{v}\cdot\vec{B}=vBcos\theta\\\\\theta=cos^{1}(\frac{\vec{v}\cdot\vec{B}}{vB})[/tex]
v: velocity of the electron
B: magnetic field
By calculating the norm of the vectors and the dot product and by replacing you obtain:
[tex]B=\sqrt{(20)^2+(50)^2+(30)^2}=61.64mT\\\\v=\sqrt{(40)^2+(30)^2+(50)^2}=70.71m/s\\\\\vec{v}\cdot\vec{B}=[(20)(40)+(50)(30)-(30)(50)]mTm/s=800mTm/s\\\\\theta=cos^{-1}(\frac{800*10^{-3}Tm/s}{(70.71m/s)(61.64*10^{-3}T)})=cos^{-1}(0.183)=1.38\°[/tex]
the angle between v and B vectors is 1.38°
b) the change in the speed of the electron can be calculated by the change in the momentum in the following way:
[tex]\frac{dp}{dt}=F_e=qvBsin\theta\\\\\frac{dp}{dt}=(1.6*10^{-19}C)(70.71m/s)(61.64*10^{-3}T)(sin(1.38\°))=6.85*10^{-19}N[/tex]
due to the mass of the electron is a constant you have:
[tex]\frac{dp}{dt}=\frac{mdv}{dt}=6.85*10^{-19}N\\\\\frac{dv}{dt}=\frac{6.85*10^{-19}N}{9.1*10^{-31}kg}=7.53*10^{11}(m/s)/s[/tex]
the change in the speed is 7.53*10^{11}m/s/s
c) the radius of the helical path is given by:
[tex]r=\frac{m_ev}{qB}=\frac{(9.1*10^{-31}kg)(70.71m/s)}{(1.6*10^{-19}C)(61.64*10^{-3}T)}=6.52*10^{-9}m[/tex]
the radius is 6.52*10^{-9}m
