An ideal solenoid having a coil density of 5000 turns per meter is 10 cm long and carries a current of 4.0
a. what is the strength of the magnetic field at its center?

The rule that is used to get the strength of magnetic field at the center of solenoid (B) is:
B = µ x n x I where:
µ is the permeability of the medium where the solenoid is based. In this problem, the medium is air which means that µ = µ o = 4 pi x 10^-7 Tm/A
I is the current passing (I = 4 amperes)
n is the number of turns per unit length (5000 turns)

Substituting in the mentioned equation, we find that:
B = 4 x 3.14 x 10^-7 x 5000 x 4 = 25.132 mT

Ke reqd to move from 0 to 50 will be proportional to 50x50 or 2500. this is equal to w.  to change vel from 50 to 150 you need energy prop to 22500 - 2500 = 20,000. this corresponds to 8w.  best answer is a) . work reqd is prop to velocity squared. 

Athermometer to measure the temperature of course