Question

A long solenoid with 1000 turns/m has a core material with relative permeability 25 and volume $10^3 cm^3$. If the core material is replaced by another material having relative permeability of 145 with same volume maintaining same current of 0.75 A in the solenoid, the fractional change in the magnetization of the core would be

Answer 1

5

Answer 2

Explanation:

1. Magnetic Intensity (H): The magnetic intensity inside a long solenoid is given by $H = nI$, where $n$ is the number of turns per unit length and $I$ is the current. Given $n = 1000 \text{ turns/m}$ and $I = 0.75 \text{ A}$. $H = 1000 \text{ turns/m} \times 0.75 \text{ A} = 750 \text{ A/m}$. Since the current and number of turns per meter remain constant, the magnetic intensity $H$ remains the same for both core materials.

2. Magnetization (M): The magnetization $M$ of a material is related to the magnetic intensity $H$ and its relative permeability $\mu_r$ by the formula: $M = (\mu_r - 1) H$

3. Initial Magnetization ($M_1$): For the first core material, $\mu_{r1} = 25$. $M_1 = (25 - 1) \times 750 = 24 \times 750 = 18000 \text{ A/m}$.

4. Final Magnetization ($M_2$): For the second core material, $\mu_{r2} = 145$. $M_2 = (145 - 1) \times 750 = 144 \times 750 = 108000 \text{ A/m}$.

5. Fractional Change in Magnetization: The fractional change is defined as $\frac{\text{Change in Magnetization}}{\text{Initial Magnetization}}$. Fractional change = $\frac{M_2 - M_1}{M_1}$ Fractional change = $\frac{108000 - 18000}{18000} = \frac{90000}{18000} = 5$.

The volume of the core material is not required for this calculation as magnetization is an intensive property (magnetic dipole moment per unit volume).