You have arranged that the magnetic field in a particular region of space is due North with a value of 0.0040 T. An electron enters the field traveling to the West with a speed of 9.5 percent of the speed of light. As a result, the electron experiences a magnetic force which is upwards. 1) What is the strength of this magnetic force?
2) What is the amount of the resulting acceleration of the electron?
3) What will be the result of the acceleration that you have calculated in part (b)?
A) The direction of the moving electron will change but its speed will remain constant.
B) Both the speed and direction of the moving electron will change.
C) The speed of the moving electron will change bu

1. Magnetic force = 4.56*10^-12 N

2. Acceleration = 5.01 *10^18 m/s^2

3. A) The direction of the moving electron will change but its speed will remain constant.

Explanation:

Given Data:

B = 0.0040 T North

q = 1.6*10^-19 C

Mass of electron = 9.1*10^-31 kg

V = 9.5% of speed of light

  = 9.5% *3*10^8

  = 28.5*10^6 m/s  west

So, B and v are perpendicular to each other.

Therefore, ∅ = 90°

1, Calculating the magnetic force using the formula;

F = qvBsin∅

  = 1.6*10^-19 * 28.5*10^6*sin90

  = 4.56*10^-12 N

Therefore, the strength of the force = 4.56*10^-12 N

2. Calculating acceleration using the formula;

F = ma

a = F/m

  = 4.56*10^-12 /9.1*10^-31

  = 5.01 *10^18 m/s^2

3. A) The direction of the moving electron will change but its speed will remain constant.

can u provide the graph and if its not possible

then i can tell u the segment where the

average time interval starts degrading is the segment where the car is slowing down

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j