Question

A very long solenoid with (2 x 2)cm² cross section has an iron core ($\mu_r$ = 1000) and 4000 turn/meter. If it carries a current of 500 mA, then

• A. Its self inductance per meter is 16$\mu_0$ × 10⁴ H/m
• B. Its self inductance per meter is 64$\mu_0$ × 10⁵ H/m
• C. The magnetic field energy stored per meter is 8$\mu_0$ × 10⁵ J/m
• D. The magnetic field energy stored per meter is 2$\mu_0$ × 10⁴ J/m

Answer 1

Answer: (B), (C)

Its self inductance per meter is 64$\mu_0$ × 10⁵ H/m

The magnetic field energy stored per meter is 8$\mu_0$ × 10⁵ J/m

Answer 2

The problem asks us to calculate the self-inductance per meter and the magnetic field energy stored per meter for a long solenoid with an iron core.

1. Given Parameters:

• Cross-sectional area, $A = (2 \text{ cm})^2 = 4 \text{ cm}^2 = 4 \times 10^{-4} \text{ m}^2$
• Relative permeability of the iron core, $\mu_r = 1000$
• Number of turns per meter, $n = 4000 \text{ turns/m}$
• Current, $I = 500 \text{ mA} = 0.5 \text{ A}$

2. Calculate Self-Inductance per Meter ($L/l$): The permeability of the core material is $\mu = \mu_r \mu_0$. So, $\mu = 1000 \mu_0$.

The formula for the self-inductance of a long solenoid is $L = \mu n^2 A l$, where $l$ is the length of the solenoid. Therefore, the self-inductance per meter is: $\frac{L}{l} = \mu n^2 A$ Substitute the given values: $\frac{L}{l} = (1000 \mu_0) \times (4000 \text{ turns/m})^2 \times (4 \times 10^{-4} \text{ m}^2)$ $\frac{L}{l} = 1000 \mu_0 \times (16 \times 10^6) \times (4 \times 10^{-4})$ $\frac{L}{l} = \mu_0 \times (10^3 \times 16 \times 10^6 \times 4 \times 10^{-4})$ $\frac{L}{l} = \mu_0 \times (16 \times 4) \times (10^{3+6-4})$ $\frac{L}{l} = \mu_0 \times 64 \times 10^5 \text{ H/m}$ $\frac{L}{l} = 64 \mu_0 \times 10^5 \text{ H/m}$ This matches option B.

3. Calculate Magnetic Field Energy Stored per Meter ($U/l$): The energy stored in an inductor is given by $U = \frac{1}{2} L I^2$. Therefore, the magnetic field energy stored per meter is: $\frac{U}{l} = \frac{1}{2} \left(\frac{L}{l}\right) I^2$ Substitute the calculated value of $L/l$ and the given current $I$: $\frac{U}{l} = \frac{1}{2} \times (64 \mu_0 \times 10^5 \text{ H/m}) \times (0.5 \text{ A})^2$ $\frac{U}{l} = \frac{1}{2} \times (64 \mu_0 \times 10^5) \times (0.25)$ $\frac{U}{l} = (32 \mu_0 \times 10^5) \times 0.25$ $\frac{U}{l} = 8 \mu_0 \times 10^5 \text{ J/m}$ This matches option C.

Both options B and C are correct.