Answer:
Explanation:
From the given information:
The missing value are as follows:
The velocity [tex]v ^{\to} = (2.0 \times 10^6 \ m/s) \hat i + (3.0 \times 10^6 \ m.s) \hat j[/tex]
The uniform magnetic field [tex]B^{\to} = (0.030 T)\hat i -(0.15 T) \hat j[/tex]
The force on electron as a result of the Magnetic field is:
[tex]F^\to = q(v^\to \times B^\to)[/tex]
here;
Change of electron [tex]q = -1.6 \times 10^{-19} C[/tex]
Then,
[tex]v^{\to} \times B^{\to} = \left|\begin{array}{ccc}2\times10^6&3\times 10^6&0\\0.03&-0.15&0\end{array}\right|[/tex]
[tex]= \hat i(0-0) -\hat j ( 0-0) +\hat k (-3\times 10^5 -0.9\times 10^5) \\ \\ =-3.9 \times 10^5 \ k[/tex]
∴
[tex]F^\to = q(v^\to \times B^\to)[/tex]
[tex]F^{\to} = -1.6 \times 10^{-19} \times -3.9 \times 10^5 \\ \\ F^{\to} = 6.24 \times 10^{-14 } \ N[/tex]
For the proton with the same velocity:
[tex]q = 1.6 \times 10^{-19 } \ C \\ \\ F = q (v^\to * B^\to) \ \\ \\ F^\to = q(-3.9 \times 10^5) \\ \\ F = (1.6 \times 10^{-19}) (-3.9 \times 10^5) \\ \\ \mathbf{F ^{\to} = -6.24 \times 10^{-14} \ N}[/tex]
