Answer:
[tex]F = (8.35 \times 10^{-16})\hat i - (12.12 \times 10^{-16})\hat j +(1.35 \times 10^{-16})\hat k[/tex]
Explanation:
When a charge is moving in constant magnetic field and electric field both then the net force on moving charge is vector sum of force due to magnetic field and electric field both
so first the force on the moving charge due to electric field is given by
[tex]\vec F_e = q\vec E[/tex]
[tex]\vec F_e = (1.6 \times 10^{-19})(3.3 \hat i - 4.5 \hat j) \times 10^3[/tex]
[tex]\vec F_e = (5.28 \times 10^{-16}) \hat i - (7.2 \times 10^{-16}) \hat j[/tex]
Now force on moving charge due to magnetic field is given as
[tex]\vec F_b = q(\vec v \times \vec B)[/tex]
[tex]\vec F_b = (1.6 \times 10^{-19})((6.6 \hat i+2.8 \hat j−4.8 \hat k) \times 10^3 \times (0.64 \hat i + 0.40 \hat j) )[/tex]
[tex]\vec F_b = (4.22 \times 10^{-16})\hat k - (2.87 \times 10^{-16})\hat k - (4.92 \times 10^{-16})\hat j + (3.07 \times 10^{-16}) \hat i[/tex]
[tex]\vec F_b = (3.07\times 10^{-16})\hat i - (4.92 \times 10^{-16})\hat j + (1.35 \times 10^{-16})\hat k[/tex]
Now net force due to both
[tex]F = F_e + F_b[/tex]
[tex]F = (8.35 \times 10^{-16})\hat i - (12.12 \times 10^{-16})\hat j +(1.35 \times 10^{-16})\hat k[/tex]