The current required to be passed through a solenoid of 15 cm length and 60 turns in order to demagnetise a bar magnet of magnetic intensity $$2.4 \times 10^3$$ A m$$^{-1}$$ is _______ A.

Given: Length $$l = 15$$ cm = 0.15 m, number of turns $$N = 60$$, magnetic intensity $$H = 2.4 \times 10^3$$ A/m.

The magnetic field intensity inside a solenoid is:

$$H = nI = \frac{N}{l}I$$

To demagnetise the bar magnet, the solenoid must produce a field of the same intensity:

$$I = \frac{Hl}{N} = \frac{2.4 \times 10^3 \times 0.15}{60} = \frac{360}{60} = 6 \text{ A}$$

The current required is 6 A.