Answer to Question 3

A current of 2.0 A flows through a coil that has 150 turns. The coil is made on a former of square section 2.0 cm × 2.0 cm, and 10 cm long. In the coil there is a material of relative permeability 150. Calculate the reluctance. 
First work out the flux density:
B = (4p × 10^{7} H m^{1} × 150 × 150 × 2.0 A) ÷ 0.10 m = 0.5655 T
Now work out flux: F = 0.5655 T × (4.0 × 10^{4} m^{2}) = 2.262 × 10^{4} Wb
Now work out the reluctance:
S = (150 × 2.0 A) ÷ 2.262 × 10^{4} Wb = 1.3 × 10^{6} H^{1} 