The coercivity of a magnet is $5 \times 10^3 \mathrm{~A} / \mathrm{m}$. The amount of current required to be passed in a solenoid of length $30 \mathrm{~cm}$ and the number of turns 150, so that the magnet gets demagnetised when inside the solenoid is ________ A.

<p>Coercivity is a measure of the resistance of a ferromagnetic material to becoming demagnetized. It is defined as the intensity of the applied magnetic field required to reduce the magnetization of a material to zero after the magnetization of the sample has been driven to saturation. In this case, coercivity $H_c$ is given to be $5 \times 10^3 \mathrm{~A/m}$.</p> <p>The relationship between the magnetic field $H$ inside a solenoid and the current $I$ passed through it is given by the formula:</p> <p>$H = \frac{N \cdot I}{L}$</p> <p>where:</p> <ul> <li>$H$ is the magnetic field strength inside the solenoid in amperes per meter (A/m),</li> <li>$N$ is the total number of turns of wire,</li> <li>$I$ is the current in amperes (A), and</li> <li>$L$ is the length of the solenoid in meters (m).</li> </ul> <p>Given that the number of turns of the solenoid $N = 150$ and the length of the solenoid $L = 30 \, \text{cm} = 0.3 \, \text{m}$, we can rearrange the formula to solve for $I$:</p> <p>$I = \frac{H \cdot L}{N}$</p> <p>Substituting the given values:</p> <p>$I = \frac{(5 \times 10^3) \times 0.3}{150}$</p> <p>Simplifying, we get:</p> <p>$I = \frac{1500}{150}$</p> <p>$I = 10 \, \text{A}$</p> <p>Therefore, the amount of current required to be passed in the solenoid for demagnetizing the magnet when inside the solenoid is <strong>10 A</strong>.</p>