Asolenoid with 3,000.0 turns is 70.0 cm long. if its self-inductance is 25.0 mh, what is 5) its radius? (the value of o is 4 × 10-7 n/a2.)

The self-inductance of a solenoid is given by:

where
is the vacuum permeability
N is the number of turns
A is the cross-sectional area of the solenoid
l is the length of the solenoid

For the solenoid in our problem, N=3000, l=70.0 cm=0.70 m and the self-inductance is L=25.0 mH=0.025 H, therefore the cross-sectional area is

And since the area is related to the radius by

The radius of the solenoid is

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j