Question 131: Magnetism and Matter

Question

The coercivity of a magnet is $$5 	imes 10^3 \mathrm{~A} / \mathrm{m}$$. The amount of current required to be passed in a solenoid of length $$30 \mathrm{~cm}$$ and the number of turns 150, so that the magnet gets demagnetised when inside the solenoid is ________ A.

Accepted Answer

Coercivity is a measure of the resistance of a ferromagnetic material to becoming demagnetized. It is defined as the intensity of the applied magnetic field required to reduce the magnetization of a material to zero after the magnetization of the sample has been driven to saturation. In this case, coercivity $$H_c$$ is given to be $$5 	imes 10^3 \mathrm{~A/m}$$.
The relationship between the magnetic field $$H$$ inside a solenoid and the current $$I$$ passed through it is given by the formula:
$$H = \frac{N \cdot I}{L}$$
where:
$$H$$ is the magnetic field strength inside the solenoid in amperes per meter (A/m),
$$N$$ is the total number of turns of wire,
$$I$$ is the current in amperes (A), and
$$L$$ is the length of the solenoid in meters (m).
Given that the number of turns of the solenoid $$N = 150$$ and the length of the solenoid $$L = 30 \, 	ext{cm} = 0.3 \, 	ext{m}$$, we can rearrange the formula to solve for $$I$$:
$$I = \frac{H \cdot L}{N}$$
Substituting the given values:
$$I = \frac{(5 	imes 10^3) 	imes 0.3}{150}$$
Simplifying, we get:
$$I = \frac{1500}{150}$$
$$I = 10 \, 	ext{A}$$
Therefore, the amount of current required to be passed in the solenoid for demagnetizing the magnet when inside the solenoid is
10 A
.

Magnetism and Matter

The coercivity of a magnet is $5 \times 10^3 \mathrm{~A} / \mathrm{m}. The amount of current required to be passed in a solenoid of length 30 \mathrm{~cm}$ and the number of turns 150, so that the magnet gets demagnetised when inside the solenoid is ________ A.

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