Question

Material of \( \mu_r = 400 \) is present inside a solenoid where the magnetic field is found to be \( 1 \, \mathrm{T} \). If the magnetic intensity here is \( \alpha \times 10^5 \) SI units, find \( \alpha \) (\( \mu_0 = 4\pi \times 10^{-7} \, \mathrm{H/m} \)).

• A. \( 1/(4\pi) \)
• B. \( 1/(16\pi) \)
• C. \( 1/(2\pi) \)
• D. \( 1/\pi \)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The magnetic field (\( B \)) inside a material is related to the magnetic intensity (\( H \)) and the permeability of the material. In a solenoid filled with a core, the total field is amplified by the relative permeability \( \mu_r \).

Step 2: Key Formula or Approach:
\[ B = \mu H = \mu_0 \mu_r H \] Where \( B \) is the magnetic induction, \( \mu_r \) is relative permeability, and \( H \) is magnetic intensity.

Step 3: Detailed Explanation:
Given \( B = 1 \, \text{T} \), \( \mu_r = 400 \), and \( \mu_0 = 4\pi \times 10^{-7} \). \[ 1 = (4\pi \times 10^{-7}) \times 400 \times H \] \[ 1 = 16\pi \times 10^{-5} \times H \] \[ H = \frac{1}{16\pi \times 10^{-5}} = \frac{10^5}{16\pi} \] We are given \( H = \alpha \times 10^5 \): \[ \alpha = \frac{1}{16\pi} \] (Note: If the question implies \( \mu_r = 4000 \) or a different constant, the value adjusts. Based on the 400 provided, the calculation leads to \( 1/16\pi \)). Re-evaluating for standard competitive question values where \( B = \mu_0 \mu_r H \) often uses \( \alpha = 2 \), let's check \( \mu_r \) again. If \( \mu_r = 400 \), then \( \alpha = 1/16\pi \).

Step 4: Final Answer:
The value of \( \alpha \) is \( \frac{1}{16\pi} \).