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 T. The magnetic intensity in SI units is $\alpha \times 10^5$. The value of $\alpha$ is ________.
(Free space permeability $\mu_0 = 4\pi \times 10^{-7}$ SI units.)
A solenoid has a core made of material with relative permeability 400. The magnetic field produced in the interior of solenoid is 1.0 T. The magnetic intensity in SI units is $\alpha \times 10^5$. The value of $\alpha$ is ________.
(Free space permeability $\mu_0 = 4\pi \times 10^{-7}$ SI units.)
(Free space permeability $\mu_0 = 4\pi \times 10^{-7}$ SI units.)
Updated On: Apr 10, 2026
- $25/\pi$
- $1/16\pi$
- $1/\pi$
- $1/4\pi$
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
The magnetic intensity $H$ (also called the magnetizing field) is related directly to the total induced magnetic field $B$ and the properties of the core material (its permeability $\mu$).
Step 2: Key Formula or Approach:
The relationship between Magnetic Field ($B$) and Magnetic Intensity ($H$):
$B = \mu H$
The permeability of the material is $\mu = \mu_0 \mu_r$, where $\mu_r$ is relative permeability.
$H = \frac{B}{\mu_0 \mu_r}$
Step 3: Detailed Explanation:
Given values:
Relative permeability, $\mu_r = 400$.
Magnetic field, $B = 1.0 \text{ T}$.
Permeability of free space, $\mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}$.
Calculate the magnetic intensity $H$:
$H = \frac{B}{\mu_0 \mu_r}$
$H = \frac{1.0}{(4\pi \times 10^{-7}) \times 400}$
$H = \frac{1}{1600\pi \times 10^{-7}}$
$H = \frac{1}{16\pi \times 10^2 \times 10^{-7}}$
$H = \frac{1}{16\pi \times 10^{-5}}$
Bringing the power of 10 to the numerator:
$H = \frac{10^5}{16\pi} \text{ A/m}$.
The problem states that the magnetic intensity is $\alpha \times 10^5$.
Equating our result to this format:
$\alpha \times 10^5 = \left(\frac{1}{16\pi}\right) \times 10^5$.
Thus, the value of $\alpha$ is $\frac{1}{16\pi}$.
Step 4: Final Answer:
The value of $\alpha$ is $1/16\pi$.
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