Question

The coercivity of a magnet is 5 × 103 A/m. The amount of current required to be passed in a solenoid of length 30 cm and the number of turns 150, so that the magnet gets demagnetised when inside the solenoid is ………A.

Answer 1

The coercivity of the magnet is given by:
$H_c = \mu_0 \frac{ni}{\mu_0},$
where:
$H_c = 5 \times 10^3 \, \text{A/m}, \quad n = \frac{\text{Number of turns}}{\text{Length of solenoid}} = \frac{150}{0.3} = 500 \, \text{turns/m}.$
Substitute into the formula:
$5 \times 10^3 = 500 \times i.$
Solve for $i$:
$i = \frac{5 \times 10^3}{500} = 10 \, \text{A}.$
Thus, the current required is:
$i = 10 \, \text{A}.$