JEE MAIN - Physics (2024 - 8th April Evening Shift - No. 29)

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.
Answer
10

Explanation

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 $$5 \times 10^3 \mathrm{~A/m}$$.

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 $$L = 30 \, \text{cm} = 0.3 \, \text{m}$$, we can rearrange the formula to solve for $$I$$:

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

Substituting the given values:

$$I = \frac{(5 \times 10^3) \times 0.3}{150}$$

Simplifying, we get:

$$I = \frac{1500}{150}$$

$$I = 10 \, \text{A}$$

Therefore, the amount of current required to be passed in the solenoid for demagnetizing the magnet when inside the solenoid is 10 A.

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