Question

An ideal solenoid with 40 turns per cm has an aluminium core and carries a current of 2.0 A. The magnetization I developed in the core and the magnetic field B at the center are:(Susceptibility of aluminium \(\chi = 2.3 \times 10^{-5}\)):

• A. \(I = 0.18 \, \text{A/m}, \, B = 3.2 \times 10^{-4} \, \text{T}\)
• B. \(I = 0.48 \, \text{A/m}, \, B = 3.2 \times 10^{-4} \, \text{T}\)
• C. \(I = 1.048 \, \text{A/m}, \, B = 3.2 \times 10^{-4} \, \text{T}\)
• D. \(I = 1.018 \, \text{A/m}, \, B = 3.2 \times 10^{-4} \, \text{T}\)

Hint:

For solenoids, always convert turns per cm to turns per meter, and include susceptibility χ when calculating the magnetization in materials.

Answer 1

The Correct Option is A

The magnetization $I$ is given by:
\[I = \chi H\]
where $H$ is the magnetic field strength. For a solenoid:
\[H = nI\]
Given $n = 40 \text{ turns/cm} = 4000 \text{ turns/m}$ and $I = 2.0 \text{ A}$:
\[H = 4000 \times 2.0 = 8000 \text{ A/m}\]
Substituting $\chi = 2.3 \times 10^{-5}$:
\[I = 2.3 \times 10^{-5} \times 8000 = 0.18 \text{ A/m}\]
The total magnetic field $B$ is:
\[B = \mu_0 (H + I), \quad \mu_0 = 4\pi \times 10^{-7} \text{ T m/A}\]
\[B = 4\pi \times 10^{-7} (8000 + 0.18) \approx 3.2 \times 10^{-4} \text{ T}\]