ExamPlay Dark Logo
وارد شوید

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.
پاسخ دهید
10

توضیح

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.

نظرات (0)

ورود به نظر
تبلیغات
BrainBehindX Inc Logo
©2026; پشتیبانی شده توسط BrainBehindX Inc