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

To determine the current required to demagnetize the magnet with coercivity \(H_c = 5 \times 10^3\) A/m, we use the magnetic field formula for a solenoid: \(B = \mu_0 nI\). Here, \(B\) represents the magnetic field, \(n\) is the turns per unit length, and \(I\) is the current.

Step 1: Calculate the turns per meter \(n\).

Given: Solenoid length \(l = 30\) cm (0.3 m), Total turns \(N = 150\).
Turns per unit length \(n = \frac{N}{l} = \frac{150}{0.3} = 500\) turns/m.

Step 2: Calculate the required current \(I\) using coercivity.

The coercivity \(H_c\) is equivalent to the magnetic field needed for demagnetization: \(H_c = nI\). Rearranging to solve for \(I\):
\(I = \frac{H_c}{n} = \frac{5 \times 10^3}{500} = 10\) A.

Step 3: Verify the calculated current against the specified range.

The calculated current \(I = 10\) A falls within the specified range of 10 to 10 A, confirming its validity.

Conclusion: The magnet requires a current of 10 A for demagnetization.