Question

A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 2.66 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2930 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) in the positive z direction and has a magnitude of 5.51 V/m, (b) in the negative z direction and has a magnitude of 5.51 V/m, and (c) in the positive x direction and has a magnitude of 5.51 V/m?

Answer 1

Answer:

a) $F = 2.13 \cdot 10^{-18} N$

b) $F = 3.65 \cdot 10^{-19} N$

c) $F = 1.53 \cdot 10^{-18} N$  

Explanation:

We have:                    

B: is the magnetic field = (2.66x10⁻³ T)(-$\hat{\imath}$)      

v: is the velocity of the proton = (2930 m/s)$\hat{\jmath}$    

a) To find the magnitude of the net force (F) we need to use the Lorentz force equation:

$\vec{F} = q(\vec{E} + \vec{v}\times \vec{B})$  (1)

E: is the electric field = (5.51 V/m)$\hat{k}$            

q: is the proton charge = 1.6x10⁻¹⁹ C

$\vec{F} = 1.6\cdot 10^{-19}(5.51 \hat{k} + 2930 \hat{\jmath}\times 2.66\cdot 10^{-3}(-\hat{\imath})) = 1.6\cdot 10^{-19}(5.51 \hat{k} + 7.79 \hat{k}) = (2.13 \cdot 10^{-18} N)\hat{k}$                                                                                            

Hence, the magnitude of the net force is:

$F = \sqrt{F_{x}^{2} + F_{y}^{2} + F_{z}^{2}} = \sqrt{0 + 0 + (2.13 \cdot 10^{-18}\hat{k})^{2}} = 2.13 \cdot 10^{-18} N$

b) When E = (5.51 V/m)($-\hat{k}$), the net force is (equation 1):

$\vec{F} = 1.6\cdot 10^{-19}[5.51(-\hat{k}) + 2930 \hat{\jmath}\times 2.66\cdot 10^{-3}(-\hat{\imath})] = 1.6\cdot 10^{-19}[5.51(-\hat{k}) + 7.79 \hat{k}] = (3.65 \cdot 10^{-19} N)\hat{k}$                     

Then, the magnitude of the net force is:

$F = \sqrt{0 + 0 + (3.65 \cdot 10^{-19}\hat{k})^{2}} = 3.65 \cdot 10^{-19} N$    

c) When E = (5.51 V/m)($\hat{\imath}$), the net force is:  

$\vec{F} = 1.6\cdot 10^{-19}(5.51\hat{\imath} + 2930 \hat{\jmath}\times 2.66\cdot 10^{-3}(-\hat{\imath})) = 1.6\cdot 10^{-19}(5.51\hat{\imath} + 7.79 \hat{k}) = 8.82 \cdot 10^{-19}\hat{\imath} + 1.25 \cdot 10^{-18}\hat{k}$                                            

Hence, the magnitude of the net force is:

$F = \sqrt{(8.82 \cdot 10^{-19}\hat{\imath})^{2} + 0 + (1.25 \cdot 10^{-18}\hat{k})^{2}} = 1.53 \cdot 10^{-18} N$  

I hope it helps you!