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Step-by-Step Solution
Step 1: Identify the relevant physical quantity
The question provides the coercivity of a small ferromagnet, which is the external magnetic field intensity required to demagnetize it. This coercivity is given as
$3 \times 10^3\, \text{A m}^{-1}$.
Step 2: Write down the expression for magnetic field in a solenoid
The magnetic field inside a long solenoid is given by
$$
B = \mu_{0} \, n \, i
$$
where
• $B$ is the magnetic field,
• $\mu_{0}$ is the permeability of free space ($= 4\pi \times 10^{-7}\,\text{N A}^{-2}$),
• $n$ is the number of turns per unit length,
• $i$ is the current through the solenoid.
Step 3: Relate coercivity to the magnetic field intensity
The coercivity value ($H_c$) is the magnetic field intensity (in
$\text{A m}^{-1}$) that must be generated by the solenoid to demagnetize the ferromagnet. In terms of the solenoid parameters:
$$
H_c = \frac{B}{\mu_0} = n i
$$
Step 4: Substitute the value of $n$ (the turn density)
The number of turns per unit length $n$ is
$$
n = \frac{N}{L}
$$
Here $N = 100$ (the total number of turns) and $L = 10\,\text{cm} = 10 \times 10^{-2}\,\text{m} = 0.1\,\text{m}$.
Thus,
$$
n = \frac{100}{0.1} = 1000\,\text{turns m}^{-1}.
$$
Step 5: Equate coercivity to $n\,i$ and solve for $i$
We have
$$
H_c = n\, i \quad \Rightarrow \quad 3 \times 10^3 = 1000 \, i.
$$
Solving for $i$,
$$
i = \frac{3 \times 10^3}{1000} = 3\,\text{A}.
$$
Step 6: Conclude the required current
Therefore, the current required to be passed in the solenoid to demagnetize the small ferromagnet (coercivity $3 \times 10^3\, \text{A m}^{-1}$) is
$3\,\text{A}$.
