191k views
3 votes
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.

User Henry C
by
8.0k points

1 Answer

2 votes

Answer:


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

Step-by-step 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 * 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 = (1.6* 10^(-19))/(1.67* 10^(-27))(57 \hat{k} - 20\hat{j})


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

User TimeEmit
by
7.5k points