Answer:
Step-by-step explanation:
Given that,
A proton moves in a magnetic field
Charge of proton q = + 1.609 × 10^-19C..
Mass of proton m = 1.67 × 10^-27 kg.
Speed of proton v = 1.8 × 10^7 m/s
Acceleration experience a = 2.2 × 10¹³ m/s²
The acceleration is in positive x-direction
Then, a = 2.2 × 10¹³ •i m/s²
The speed moves along positive direction
Then, v = 1.8 × 10^7 •k m/s
Magnetic Field and it's direction?
The force that cause the proton to be in motion in the magnetic field is given as
F = qvBSinθ
Making B subject of formulas
B = F / qvSinθ
The velocity and the field are perpendicular, then θ=90°
Also, from newton second law, F=ma
Then,
B = ma / qvSinθ
Sin90 = 1
So,
B = 1.67 × 10^-27 × 2.2 × 10¹³ / 1.609 × 10^-19 × 1.8 × 10^7
B = 0.0127 T
B = 1.27 × 10^-2 T
Direction
Using the right-hand thumb rule, since the acceleration is in the positive x-direction and the velocity is in the positive z-direction, then, the magnetic field must be in the negative y-direction
Prove
F = q (V × B). Cross product of V and B
m(a) •i = q(V •k × B )
We know that, j × k = i and k × j = -i,
Since, k × j = -i, then k × -j = i
So, B must be in the negative y direction
Therefore,
B = —1.27 × 10^-2 •j T