The coercivity of a magnet is $$5 \times 10^3$$ 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.

We need to find the current required to demagnetise a magnet placed inside a solenoid, given the coercivity of the magnet.

Coercivity is the intensity of the magnetising field ($$H$$) required to reduce the magnetisation of a material to zero after it has been magnetised. In other words, to demagnetise the magnet, we need to apply a magnetic field intensity equal to its coercivity, which is $$H = 5 \times 10^3$$ A/m.

Since the magnetic field intensity inside a solenoid is given by $$H = nI = \frac{N}{L} \times I$$, where $$N$$ is the total number of turns, $$L$$ is the length of the solenoid, and $$I$$ is the current, we can use this relation to determine the required current.

Next, substituting $$N = 150$$ turns, $$L = 30$$ cm $$= 0.3$$ m, and $$H = 5 \times 10^3$$ A/m into the expression gives $$5 \times 10^3 = \frac{150}{0.3} \times I$$, which simplifies to $$5 \times 10^3 = 500 \times I$$. From this, $$I = \frac{5 \times 10^3}{500} = 10 \text{ A}$$.

The answer is 10 A.