Explanation:
It is given that,
Velocity of the electron, 
Magnetic field, 
Charge of electron, 
(a) Let 
is the force on the electron due to the magnetic field. The magnetic force acting on it is given by :
![F_e=1.6\times 10^{-19}\times [(2\times 10^6i+3\times 10^6j)\times (0.030i-0.15j)]](https://tex.z-dn.net/?f=F_e%3D1.6%5Ctimes%2010%5E%7B-19%7D%5Ctimes%20%5B%282%5Ctimes%2010%5E6i%2B3%5Ctimes%2010%5E6j%29%5Ctimes%20%280.030i-0.15j%29%5D)
(b) The charge of electron, 
The force acting on the proton is same as force on electron but in opposite direction i.e (-k). Hence, this is the required solution.
Answer:
The object will travel at a constant rate in along a straight line.
Explanation:
In the given situation, it is mentioned that there is no external force acting on the given object. Thus, it will retain its initial velocity along a straight path.
Answer:
(a) 0.613 m
(b) 0.385 m
(c) vₓ = 1.10 m/s, vᵧ = 3.50 m/s
v = 3.68 m/s², θ = 72.6° below the horizontal
Explanation:
(a) Take down to be positive.
Given in the y direction:
v₀ = 0 m/s
a = 10 m/s²
t = 0.350 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (0 m/s) (0.350 s) + ½ (10 m/s²) (0.350 s)²
Δy = 0.613 m
(b) Given in the x direction:
v₀ = 1.10 m/s
a = 0 m/s²
t = 0.350 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (1.10 m/s) (0.350 s) + ½ (0 m/s²) (0.350 s)²
Δx = 0.385 m
(c) Find: vₓ and vᵧ
vₓ = aₓt + v₀ₓ
vₓ = (0 m/s²) (0.350 s) + 1.10 m/s
vₓ = 1.10 m/s
vᵧ = aᵧt + v₀ᵧ
vᵧ = (10 m/s²) (0.350 s) + 0 m/s
vᵧ = 3.50 m/s
The magnitude is:
v² = vₓ² + vᵧ²
v = 3.68 m/s²
The direction is:
θ = atan(vᵧ / vₓ)
θ = 72.6° below the horizontal
