Answer:
The answer is "
"
Explanation:
The magnetic field at ehe mid point of the coils is,

Here, i is the current through the loop, R is the radius of the loop and x is the distance of the midpoint from the loop.


Calculating the force experienced through the protons:

Answer: 10 m/s
We're told the speed is constant, so it's not changing throughout the time period given to us. So throughout the entire interval, the speed is 10 m/s.
Answer:
A
Explanation:
The line(A) goes throughout the entire picture. So therefore choice A would be it's length.
We are given with the expression d = ut + 0.5 at^2 and is asked to express the equation in terms of a. First, we transpose ut to the left side, then we multiply to the equation and divide lastly the resulting equation by t^2. The final expression becomes a = 2(d-ut)/t^2.