Answer:
a) [tex] F = 2.13 \cdot 10^{-18} N [/tex]
b) [tex] F = 3.65 \cdot 10^{-19} N [/tex]
c) [tex] F = 1.53 \cdot 10^{-18} N [/tex]
Explanation:
We have:
B: is the magnetic field = (2.66x10⁻³ T)(-[tex]\hat{\imath}[/tex])
v: is the velocity of the proton = (2930 m/s)[tex]\hat{\jmath}[/tex]
a) To find the magnitude of the net force (F) we need to use the Lorentz force equation:
[tex] \vec{F} = q(\vec{E} + \vec{v}\times \vec{B}) [/tex] (1)
E: is the electric field = (5.51 V/m)[tex]\hat{k}[/tex]
q: is the proton charge = 1.6x10⁻¹⁹ C
[tex]\vec{F} = 1.6\cdot 10^{-19}(5.51 \hat{k} + 2930 \hat{\jmath}\times 2.66\cdot 10^{-3}(-\hat{\imath})) = 1.6\cdot 10^{-19}(5.51 \hat{k} + 7.79 \hat{k}) = (2.13 \cdot 10^{-18} N)\hat{k}[/tex]
Hence, the magnitude of the net force is:
[tex]F = \sqrt{F_{x}^{2} + F_{y}^{2} + F_{z}^{2}} = \sqrt{0 + 0 + (2.13 \cdot 10^{-18}\hat{k})^{2}} = 2.13 \cdot 10^{-18} N[/tex]
b) When E = (5.51 V/m)([tex]-\hat{k}[/tex]), the net force is (equation 1):
[tex]\vec{F} = 1.6\cdot 10^{-19}[5.51(-\hat{k}) + 2930 \hat{\jmath}\times 2.66\cdot 10^{-3}(-\hat{\imath})] = 1.6\cdot 10^{-19}[5.51(-\hat{k}) + 7.79 \hat{k}] = (3.65 \cdot 10^{-19} N)\hat{k}[/tex]
Then, the magnitude of the net force is:
[tex]F = \sqrt{0 + 0 + (3.65 \cdot 10^{-19}\hat{k})^{2}} = 3.65 \cdot 10^{-19} N[/tex]
c) When E = (5.51 V/m)([tex]\hat{\imath}[/tex]), the net force is:
[tex]\vec{F} = 1.6\cdot 10^{-19}(5.51\hat{\imath} + 2930 \hat{\jmath}\times 2.66\cdot 10^{-3}(-\hat{\imath})) = 1.6\cdot 10^{-19}(5.51\hat{\imath} + 7.79 \hat{k}) = 8.82 \cdot 10^{-19}\hat{\imath} + 1.25 \cdot 10^{-18}\hat{k}[/tex]
Hence, the magnitude of the net force is:
[tex]F = \sqrt{(8.82 \cdot 10^{-19}\hat{\imath})^{2} + 0 + (1.25 \cdot 10^{-18}\hat{k})^{2}} = 1.53 \cdot 10^{-18} N[/tex]
I hope it helps you!
