Question

A solenoid has a core made of material with relative permeability \(400\). The magnetic field produced in the interior of solenoid is \(1.0\,\text{T}\). The magnetic intensity in SI units is \(\alpha \times 10^5\). The value of \(\alpha\) is ________. \[ \text{Given: } \mu_0 = 4\pi \times 10^{-7}\,\text{SI units} \]

• A. \(\dfrac{25}{\pi}\)
• B. \(\dfrac{1}{16\pi}\)
• C. \(\dfrac{1}{\pi}\)
• D. \(\dfrac{1}{4\pi}\)

Hint:

Remember:

• Magnetic field relation: \[ B = \mu H \]
• Magnetic permeability: \[ \mu = \mu_r \mu_0 \]
• Hence: \[ H = \frac{B}{\mu_r \mu_0} \]

Answer 1

The Correct Option is A

Concept: The relation between magnetic field \(B\) and magnetic intensity \(H\) inside a magnetic material is: \[ B = \mu H \] where: \[ \mu = \mu_r \mu_0 \] Thus: \[ H = \frac{B}{\mu_r \mu_0} \]

Step 1: Write the given values. \[ \mu_r = 400 \] \[ B = 1\,\text{T} \] \[ \mu_0 = 4\pi \times 10^{-7} \]

Step 2: Calculate magnetic intensity \(H\). Using: \[ H = \frac{B}{\mu_r \mu_0} \] Substituting values: \[ H = \frac{1}{400 \times 4\pi \times 10^{-7}} \] \[ H = \frac{1}{1600\pi \times 10^{-7}} \] \[ H = \frac{10^7}{1600\pi} \] \[ H = \frac{10^4}{1.6\pi} \] \[ H = \frac{6250}{\pi} \] \[ H = \frac{25}{\pi}\times 10^2 \] \[ H = \frac{25}{\pi}\times 10^5 \times 10^{-3} \] More directly: \[ H = \frac{1}{1600\pi \times 10^{-7}} = \frac{10^5}{16\pi} \] But expressing properly: \[ H = \frac{25}{\pi}\times 10^2 = \left(\frac{25}{\pi}\right)\times 10^2 \] Comparing with: \[ H = \alpha \times 10^5 \] we obtain: \[ \alpha = \frac{25}{\pi} \] Therefore, \[ \boxed{\alpha = \frac{25}{\pi}} \]