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Worked Examples · Example 5.5

Q.A solenoid has a core of a material with relative permeability 400400. The windings of the solenoid are insulated from the core and carry a current of 2 A2\ \text{A}. If the number of turns is 10001000 per metre, calculate

(a) HH,
(b) MM,
(c) BB and
(d) the magnetising current ImI_m.
Puducherry CbseNCERTSubjective· 3mImportance★★★★★
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Using the magnetic field intensity HH from the solenoid current, we find B=μrμ0HB = \mu_r \mu_0 H, then magnetization M=(μr−1)HM = (\mu_r - 1)H, and the magnetising current Im=M/nI_m = M / n. The results are H=2000 A/mH = 2000\ \text{A/m}, M=7.98×105 A/mM = 7.98 \times 10^5\ \text{A/m}, B=1.005 TB = 1.005\ \text{T}, and Im=798 AI_m = 798\ \text{A}.

The core of this problem is understanding the three magnetic quantities — HH, MM, and BB — and how they relate inside a material. In a solenoid, the field produced by the free current alone is HH. The material responds by developing a magnetization MM, which adds to HH to give the total magnetic field BB. The relative permeability μr\mu_r tells us how strongly the material amplifies the field.

Let’s work through each part step by step.

  1. Find HH — the magnetic field intensity For a long solenoid, HH depends only on the free current and the number of turns per metre, not on the core material.

H=nIH = n I

where n=1000 turns/mn = 1000\ \text{turns/m} and I=2 AI = 2\ \text{A}.

H=1000×2=2000 A/mH = 1000 \times 2 = 2000\ \text{A/m}

This is the field that would exist in vacuum if the core were absent.

  1. Find BB — the magnetic flux density Inside a linear magnetic material, BB is related to HH by:

B=μH=μrμ0HB = \mu H = \mu_r \mu_0 H

Given μr=400\mu_r = 400 and μ0=4π×10−7 H/m\mu_0 = 4\pi \times 10^{-7}\ \text{H/m}:

B=400×(4π×10−7)×2000B = 400 \times (4\pi \times 10^{-7}) \times 2000

B=400×8π×10−4=3200π×10−4B = 400 \times 8\pi \times 10^{-4} = 3200\pi \times 10^{-4}

B=1.0053 T≈1.005 TB = 1.0053\ \text{T} \approx 1.005\ \text{T}

Watch out

A common mistake is to forget that BB uses μ0\mu_0 times μr\mu_r, not just μr\mu_r times HH as a number. Always include μ0=4π×10−7\mu_0 = 4\pi \times 10^{-7}.

  1. Find MM — the magnetization Magnetization MM is the magnetic moment per unit volume of the core material. The fundamental relation is:

B=μ0(H+M)B = \mu_0 (H + M)

Rearranging:

M=Bμ0−HM = \frac{B}{\mu_0} - H

Substitute BB from step 2:

M=1.00534π×10−7−2000M = \frac{1.0053}{4\pi \times 10^{-7}} - 2000

First term: 1.00531.2566×10−6≈8.00×105 A/m\frac{1.0053}{1.2566 \times 10^{-6}} \approx 8.00 \times 10^5\ \text{A/m}

So:

M=8.00×105−2000=7.98×105 A/mM = 8.00 \times 10^5 - 2000 = 7.98 \times 10^5\ \text{A/m}

Alternatively, using μr\mu_r directly: …

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