To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.
This definition is described in the following equation as,

Where,
permeability of free space
Number of turns in solenoid 1
Number of turns in solenoid 2
Cross sectional area of solenoid
l = Length of the solenoid
Part A )
Our values are given as,





Substituting,



PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.
Explanation:
The compass needle moved when the wire was connected to the battery. The important point here is that the needle is affected by the wire only when both ends of the wire are connected to the battery because only at this time is current flowing through the circuit.
Answer
given,
diameter of the pipe is = (14 ft)4.27 m
minimum speed of the skater must have at very top = ?
At the topmost point of the pipe the normal force will be equal to zero.
F = mg
centripetal force acting on the skateboard

equating both the force equation


r = d/2 = 14/ 2 = 7 ft
or
r = 4.27/2 = 2.135 m
g = 32 ft/s² or g = 9.8 m/s²

v = 14.96 ft/s
or

v = 4.57 m/s
Answer: 
Explanation:
This problem can be solved by the following equation:

Where:
is the change in kinetic energy
is the electric potential difference
is the electric charge
Finding
:


Finally:
