The current required to be passed through a solenoid of 15 cm length and 60 turns in order of demagnetise a bar magnet of magnetic intensity $2.4\times10^3~Am^{-1}$ is ___________ A.

<p>We know that the magnetizing field (H) inside a solenoid is given by the formula :</p> <p>$ H = \frac{N \cdot I}{L} $</p> <p>where (N) is the number of turns, (I) is the current in Amperes, and (L) is the length of the solenoid in meters.</p> <p>To demagnetize a bar magnet that has a magnetic intensity (H) of ($2.4 \times 10^3 \, \text{Am}^{-1}$), you need to create a magnetizing field in the solenoid that is equal in magnitude but opposite in direction.</p> <p>Given:</p> <ul> <li>($H = 2.4 \times 10^3 \, \text{Am}^{-1}$)</li> <li>(N = 60) turns</li> <li>($L = 15 \, \text{cm} = 0.15 \, \text{m}$) (since $(1\, \text{m} = 100 \, \text{cm})$)</li> </ul> <p>You can rearrange the formula for (H) to solve for (I) :</p> <p>$ I = \frac{H \cdot L}{N} $</p> <p>Substituting the given values, you get :</p> <p>$ I = \frac{(2.4 \times 10^3 \, \text{Am}^{-1}) \cdot 0.15 \, \text{m}}{60} $</p> <p>$ I = 6 \, \text{Amperes} $</p> <p>So, the current required to be passed through the solenoid to demagnetize the bar magnet is (6 $\, \text{A}$).</p>