We know that the magnetizing field (H) inside a solenoid is given by the formula :

$ H = \frac{N \cdot I}{L} $

where (N) is the number of turns, (I) is the current in Amperes, and (L) is the length of the solenoid in meters.

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.

Given:

• ($H = 2.4 \times 10^3 \, \text{Am}^{-1}$)
• (N = 60) turns
• ($L = 15 \, \text{cm} = 0.15 \, \text{m}$) (since $(1\, \text{m} = 100 \, \text{cm})$)

You can rearrange the formula for (H) to solve for (I) :

$ I = \frac{H \cdot L}{N} $

Substituting the given values, you get :

$ I = \frac{(2.4 \times 10^3 \, \text{Am}^{-1}) \cdot 0.15 \, \text{m}}{60} $

$ I = 6 \, \text{Amperes} $

So, the current required to be passed through the solenoid to demagnetize the bar magnet is (6 $\, \text{A}$).