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PhyMedMain 2026 • 4th Apr - Shift 2

Determine Magnetic Intensity in Solenoid with Ferromagnetic Core

A solenoid has a core made of material with relative permeability 400400. The magnetic field produced in the interior of solenoid is 1.0 T1.0\text{ T}. The magnetic intensity in SI units is α×105\alpha \times 10^5. The value of α\alpha is ‾\underline{\quad\quad\quad}.

(Free space permeability μ0=4π×10−7 SI units.\mu_0 = 4\pi \times 10^{-7}\text{ SI units.})

Options

A

25π\frac{25}{\pi}

B

116π\frac{1}{16\pi}

Correct
C

1π\frac{1}{\pi}

D

14π\frac{1}{4\pi}

Topics & Concepts

Magnetism and MatterMagnetic Force

Step-by-Step Solution

To find the value of α\alpha, we start with the relationship between the total magnetic field (BB), the relative permeability of the core material (μr\mu_r), the permeability of free space (μ0\mu_0), and the magnetic intensity (HH):

B=μH=μrμ0HB = \mu H = \mu_r \mu_0 H

Rearranging the formula to solve for the magnetic intensity HH:

H=Bμrμ0H = \frac{B}{\mu_r \mu_0}

Given parameters:

  • Magnetic field, B=1.0 TB = 1.0 \text{ T}
  • Relative permeability, μr=400\mu_r = 400
  • Permeability of free space, μ0=4π×10−7 SI units\mu_0 = 4\pi \times 10^{-7} \text{ SI units}

Substituting these values into the equation:

H=1.0400×(4π×10−7)H = \frac{1.0}{400 \times (4\pi \times 10^{-7})}

H=1.01600π×10−7H = \frac{1.0}{1600\pi \times 10^{-7}}

H=116π×10−5H = \frac{1}{16\pi \times 10^{-5}}

H=(116π)×105 A/mH = \left(\frac{1}{16\pi}\right) \times 10^5 \text{ A/m}

It is given that the magnetic intensity in SI units is α×105\alpha \times 10^5. Comparing the two expressions:

α×105=(116π)×105\alpha \times 10^5 = \left(\frac{1}{16\pi}\right) \times 10^5

α=116π\alpha = \frac{1}{16\pi}

Thus, the correct option is B.