Answer
$I = 11.67 A$
Work Step by Step
Given:
iron bar with coercivity of 7000 A/m
cylindrical wire coil 0.25 m long with 150 turns
Required:
electric current to generate the necessary magnetic field
Solution:
To demagnetize a magnet with coercivity of 7000 A/m, an H field of 7000 A/m must be applied in the opposite direction of magnetization. Using Equation 20.1:
$I = \frac{HI}{N} = \frac{(7000 A/m)(0.25 m)}{150 turns} = 11.67 A$
