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

Correct Answer: 10

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:
\[i = 10 \, \text{A}.\]