Asolenoid used to produce magnetic fields for research purposes is 2.4 m long, with an inner radius of 30 cm and 1500 turns of wire. when running, the solenoid produced a field of 1.1 t in the center. part a given this, how large a current does it carry?

= 1396.83 Amperes

For a solenoid, B = u0 * N/L * I

u0 = 1.26 * 10^-6

B = 1.1 tesla

N/L = number of turns ÷ length of the solenoid

      = 1500 ÷ 2.4

1.1 = 1.26 * 10^-6 * (1500 ÷ 2.4) * I

I = 1.1 ÷ [1.26 * 10^-6 * (1500 ÷ 2.4)]

I = 1396.83 Amperes

a lot of it means it can be "burned" for energy and sold at one price. if it is scarce, the price will go up

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray