Question

A proton moves through a region containing a uniform electric field given by E = 56.0 j V/m and a uniform magnetic field B = (0.200i + 0.300j + 0.400k ) T. Determine the acceleration of the proton when it has a velocity v = 190i m/s.

Answer 1

Answer:

$a = \dfrac{(9.57\times 10^{7})(57 \hat{k} - 20\hat{j})}$

Explanation:

given,

E = 56.0 j V/m

B = (0.2 i + 0.3 j + 0.4 k ) T

v = 190 i m/s

mass of proton = 1.67 x 10⁻²⁷ Kg

e = 1.6 x 10⁻¹⁹ C

acceleration of proton is equal to = ?

magnetic force

F_B = e v B

F_B = e x (190 i) x (0.2 i + 0.3 j + 0.4 k)

F_B = e x (57 k - 76 j)

for proton electric force

$F_e = 56 \times e \hat{j}$

$F_{net} = F_A + F_B$

$F_{net} = e(57 \hat{k} - (76-56) \hat{j})$

$F_{net} = e(57 \hat{k} - 20\hat{j})$

$a = \dfrac{1.6\times 10^{-19}}{1.67\times 10^{-27}}(57 \hat{k} - 20\hat{j})$

$a = (9.57\times 10^{7})(57 \hat{k} - 20\hat{j})$