Q.A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2 A. If the number of turns is 1000 per metre, calculate
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Magnetic Materials Magnetization
From a Paperclip to a Magnet: The Intuition
You already know that a magnet can pick up iron nails. But what is actually happening inside that nail when it gets near the magnet? And why does a plastic comb, rubbed on hair, pick up tiny bits of paper — but never iron filings?
The answer lies in magnetization — the process by which a material becomes magnetic.
Think of a piece of iron as a chaotic crowd of tiny compass needles. Each needle is an atomic magnetic moment (a tiny magnet, arising from the spin of electrons). In unmagnetized iron, these needles point in random directions. Their magnetic effects cancel out, so the iron as a whole shows no net magnetism.
Now bring a strong magnet close. Its magnetic field acts like a command: "Line up!" The tiny compass needles inside the iron start rotating, aligning themselves with the external field. The more they align, the stronger the iron's own magnetic field becomes. This alignment is magnetization.
Magnetization is not the same as inducing a current. It is a purely magnetic reorientation of atomic dipoles inside a material.
The Precise Definition
Magnetization (M) is the net magnetic dipole moment per unit volume of a material. It tells you how strongly a material is magnetized — how many tiny atomic magnets are aligned, and in which direction.
If a material has N atoms per unit volume, each with an average magnetic moment μavg, then:
M=Nμavg
The SI unit of M is amperes per metre (A/m). Why? Because a magnetic dipole moment has units of A·m², and dividing by volume (m³) gives A/m.
M=volumetotal magnetic dipole moment
How Magnetization Connects to the Magnetic Field
When a material gets magnetized, it produces its own magnetic field. The total magnetic field B inside the material is the sum of:
- The external applied field H (caused by free currents, like the current in a solenoid)
- The material's response — the magnetization M
The fundamental relation is:
B=μ0(H+M)
where μ0=4π×10−7T⋅m/A is the permeability of free space.
Do not confuse H (magnetic field intensity, or "magnetizing field") with B (magnetic flux density). H is what you apply; M is what the material does; B is the total field you measure.
The Three Kinds of Magnetic Materials
Not all materials respond the same way to an external field. The magnetization M is proportional to H for most materials (at least for small fields):
M=χmH
where χm is the magnetic susceptibility — a dimensionless number that tells you how easily a material magnetizes.
| Material Type | χm | Behaviour | Example |
|---|---|---|---|
| Diamagnetic | Small and negative (≈−10−5) | Weakly repelled by a magnet; M opposes H | Water, copper, bismuth |
| Paramagnetic | Small and positive (≈10−5 to 10−3) | Weakly attracted; M aligns with H | Aluminium, oxygen gas |
| Ferromagnetic | Large and positive (≫1) | Strongly attracted; M can be huge and persists even after H is removed | Iron, nickel, cobalt |
Why this formula?
Magnetic Materials & Magnetization: Why the Key Formulas Hold
Let's build this from the ground up — starting with what magnetization physically means, then deriving the formulas step by step.
1. What is Magnetization (M)?
Magnetization is the net magnetic dipole moment per unit volume of a material.
- Inside a material, atoms act like tiny magnetic dipoles (due to electron spin and orbital motion).
- Without an external field, these dipoles point randomly → net M=0.
- When an external field H is applied, dipoles align partially → net M=0.
Definition:
M=volumenet magnetic dipole moment
Units: A/m (same as H).
2. The Fundamental Relation: B=μ0(H+M)
This is the master equation linking the three magnetic fields:
- B = magnetic flux density (the total field inside the material)
- H = applied magnetic field (due to free currents)
- M = magnetization (response of the material)
- μ0 = permeability of free space (4π×10−7 H/m)
Why this form?
Step 1: In vacuum, there is no material, so M=0. Then:
B=μ0H
Step 2: Inside a material, the dipoles themselves produce an additional field. The total B is the sum of:
- The field due to free currents (μ0H)
- The field due to bound currents (from aligned dipoles), which is μ0M
Hence:
B=μ0H+μ0M=μ0(H+M)
Key insight: M is not an independent field — it's the material's response to H.
3. Magnetic Susceptibility (χm) and Permeability (μ)
For linear, isotropic, homogeneous materials (most common in exams), magnetization is proportional to the applied field:
M=χmH
- χm = magnetic susceptibility (dimensionless)
- χm>0 for paramagnetic materials
- χm<0 for diamagnetic materials
- χm≫1 for ferromagnetic materials (but not linear!)
Derivation of relative permeability μr:
Substitute M=χmH into the master equation:
B=μ0(H+χmH)=μ0(1+χm)H
Define:
μr=1+χm(relative permeability)
μ=μ0μr(absolute permeability)
Thus:
B=μH
Why this matters: It shows that the material simply scales the applied field by a factor μr.
4. Why χm Has Different Signs (Physical Reasoning)
| Material Type | χm | Why? |
|---|---|---|
| Diamagnetic | χm<0 (small, ~10−5) | Applied field induces opposing dipole moments (Lenz's law at atomic level). M opposes H. |
| Paramagnetic | χm>0 (small, ~10−3) | Permanent atomic dipoles align partially with H. Thermal agitation fights alignment. |
Concept: Magnetic Materials – Magnetization. The magnetic field H is due to the free current in the solenoid; the material’s response adds magnetization M, giving the total field B=μ0(H+M). Relative permeability μr relates B and H via B=μrμ0H.
Step 1 – Magnetizing field H
For an ideal solenoid, H=nI, where n=1000 turns/m and I=2 A.
H=1000×2=2000 A/m
Step 2 – Magnetic flux density B
Using B=μrμ0H with μr=400 and μ0=4π×10−7 H/m:
B=400×(4π×10−7)×2000=1.0053 T(approx 1.01 T)
Step 3 – Magnetization M
From B=μ0(H+M), we get M=μ0B−H. …
Using the magnetic field intensity H from the solenoid current, we find B=μrμ0H, then magnetization M=(μr−1)H, and the magnetising current Im=M/n. The results are H=2000 A/m, M=7.98×105 A/m, B=1.005 T, and Im=798 A.
The core of this problem is understanding the three magnetic quantities — H, M, and B — and how they relate inside a material. In a solenoid, the field produced by the free current alone is H. The material responds by developing a magnetization M, which adds to H to give the total magnetic field B. The relative permeability μr tells us how strongly the material amplifies the field.
Let’s work through each part step by step.
- Find H — the magnetic field intensity For a long solenoid, H depends only on the free current and the number of turns per metre, not on the core material.
H=nI
where n=1000 turns/m and I=2 A.
H=1000×2=2000 A/m
This is the field that would exist in vacuum if the core were absent.
- Find B — the magnetic flux density Inside a linear magnetic material, B is related to H by:
B=μH=μrμ0H
Given μr=400 and μ0=4π×10−7 H/m:
B=400×(4π×10−7)×2000
B=400×8π×10−4=3200π×10−4
B=1.0053 T≈1.005 T
A common mistake is to forget that B uses μ0 times μr, not just μr times H as a number. Always include μ0=4π×10−7.
- Find M — the magnetization Magnetization M is the magnetic moment per unit volume of the core material. The fundamental relation is:
B=μ0(H+M)
Rearranging:
M=μ0B−H
Substitute B from step 2:
M=4π×10−71.0053−2000
First term: 1.2566×10−61.0053≈8.00×105 A/m
So:
M=8.00×105−2000=7.98×105 A/m
Alternatively, using μr directly: …
Method: Magnetization Relations in a Solenoid Core
We use the magnetic circuit approach with definitions of magnetizing field (H), magnetization (M), and magnetic flux density (B).
Step 1: Calculate H (Magnetizing Field)
H depends only on the free current in the solenoid windings — not on the core material.
H=nI
where:
- n=1000 turns/m
- I=2 A
H=1000×2=2000 A/m
Step 2: Calculate B (Magnetic Flux Density)
Using the relation with relative permeability μr=400:
B=μ0μrH
where μ0=4π×10−7 H/m
B=(4π×10−7)×400×2000
B=4π×10−7×8×105
B=3.2π×10−1=1.0053 T (approximately 1.01 T)
Step 3: Calculate M (Magnetization)
From the fundamental relation:
B=μ0(H+M)
Rearrange:
M=μ0B−H
M=4π×10−71.0053−2000
M=8×105−2000=7.98×105 A/m
Alternatively, using M=(μr−1)H:
M=(400−1)×2000=399×2000=7.98×105 A/m
Step 4: Calculate Im (Magnetising Current)
Magnetising current is the equivalent current that would produce the same B if the core were absent. It is related to M by:
Im=M×length per turn? …
Here are the common mistakes students make on this problem, along with how to avoid each.
1. Confusing H with B
Mistake:
Students often plug μr into the formula for H, writing H=μrnI.
Why it’s wrong:
H (magnetizing field) depends only on the free current and geometry — not on the core material.
The formula is:
H=nI
where n=1000 turns/m and I=2 A.
How to avoid:
Remember: H is the field due to free currents alone. The core’s response comes later, in B and M.
2. Forgetting to convert units or misreading n
Mistake:
Using n=1000 without checking units, or writing n=1000 turns instead of 1000 turns/m.
How to avoid:
Always write units explicitly. Here n=1000 m−1 is already given. If it were “1000 turns per metre,” that’s exactly n=1000.
3. Using B=μ0H for a magnetic core
Mistake:
Writing B=μ0H even when a core is present.
Why it’s wrong:
Inside a material, B=μ0μrH. For air, μr=1, but here μr=400.
Correct formula:
B=μ0μrH
How to avoid:
Always check: is there a core? If yes, use μ=μ0μr.
4. Mixing up M and B or M and H
Mistake:
Writing M=χmH but forgetting χm=μr−1, or writing M=μrH.
Correct relation:
M=χmH=(μr−1)H
How to avoid:
Memorize the chain:
μr→χm=μr−1→M=χmH
5. Sign errors in the magnetizing current Im
Mistake:
Writing Im=(μr−1)I without considering n.
Why it’s wrong:
Magnetizing current is defined as the equivalent current that would produce the same B if the core were absent. The correct formula is:
Im=(μr−1)I
Standard definition:
Im=(μr−1)I
where I is the free current. However, some textbooks define it as:
Im=(μr−1)nI(per unit length)
How to avoid:
Check your textbook’s definition. In most Indian board exams (CBSE, NCERT pattern), the magnetizing current per unit length is:
Im=(μr−1)I …
- GUJCET 2026Set x1 markMCQQ.A closely wound solenoid of 800 turns and area of cross section 2.5×10−4 m2 can carry a current of 3.0 A. The magnetic moment associated with it is ______. (A) 60 JT−1 (B) 0.60 JT−1 (C) 6 JT−1 (D) 0.06 JT−1
›Reveal solutionSolution
m=NIA=0.60 J T−1.
Magnetic moment of a solenoid: …
- GUJCET 2024Set 131 markMCQQ.AmVs is the unit of which physical quantity? (A) χm (B) μ0 (C) χc (D) ε0
›Reveal solutionSolution
A⋅mV⋅s equals AT⋅m, which is the SI unit of the permeability of free space μ0 (H/m).
Concept. μ0 has units of henry per metre; 1H=1V⋅s/A, so μ0 is measured in A⋅mV⋅s.
Steps. …
- GUJCET 2024Set 131 markMCQQ.A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2 A. If the number of turns is 1000 per meter then the value of magnetic intensity will be ________. (A) 8×10−5 Am−1 (B) 2×103 Am−1 (C) 2×10−3 Am−1 (D) 8×105 Am−1
›Reveal solutionSolution
H=nI — it depends only on the current and turns per metre, not on the core.
Concept. Magnetic intensity (magnetising field) in a solenoid is H=nI, independent of the core material's permeability. …
- GUJCET 2023Set 091 markMCQQ.A solenoid of length 0.5 m has a radius of 1 cm and is made up of 250 turns. It carries a current of 5 A. What is the magnitude of the magnetic field inside the solenoid? (A) 3.14×10−3 T (B) 6.28×10−3 T (C) 62.8×10−3 T (D) Zero
›Reveal solutionSolution
Field inside a long solenoid is B=μ0nI; the radius is irrelevant.
Concept: For a solenoid, B=μ0nI where n is turns per unit length.
n=LN=0.5250=500 m−1. …
- GUJCET 2023Set 091 markMCQQ.A solenoid has a core of a material with relative permeability 400. The windings of solenoid are insulated from the core and carry a current of 2 A. If the number of turns is 1000 per metre, the magnetic field B inside the solenoid is ______ T. (A) 1.5 (B) 1.0 (C) 1.8 (D) 2.0
›Reveal solutionSolution
A magnetic core multiplies the solenoid field by relative permeability: B=μ0μrnI.
Concept: With a core, B=μ0μrnI. …
- GUJCET 2022Set 171 markMCQQ.A solenoid of length 0.25 m has a radius of 1 cm and is made up of 500 turns. It carries a current of 2.5 A. What is the magnitude of the magnetic field inside the solenoid? (μ0=4π×10−7 SI) (A) 6.28×10−3 T (B) 6.28×10−2 T (C) 6.28×10−4 T (D) 6.28×10−1 T
›Reveal solutionSolution
n=N/l=2000 turns/m, so B=μ0nI=6.28×10−3 T (radius irrelevant).
Concept: n=0.25500=2000 turns/m. …
- GUJCET 2022Set 171 markMCQQ.A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 1 A. If the number of turns is 1000 per metre, find magnetic field (B) ________ T. (μ0=4π×10−7 SI) (A) 1.6π×10+2 (B) 16π×102 (C) 16π×10−2 (D) 0.16π×10−2
›Reveal solutionSolution
With a magnetic core, B=μ0μrnI=16π×10−2 T. …
- GUJCET 2021Set 151 markMCQQ.A solenoid of length 0.5 m has a radius of 1 cm and is made up of 1000 turns. It carries a current of 10A. What is the magnitude of the magnetic field inside the solenoid? (A) 6.28×10−3 T (B) 2.51×10−2 T (C) 1.71×10−2 T (D) 7.23×10−3 T
›Reveal solutionSolution
B=μ0nI; radius is irrelevant for a long solenoid.
Concept: …
- GUJCET 2020Set 071 markMCQQ.The relative permeability in a core of a solenoid is 400. The windings of a solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per meter. Then magnetic Intensity inside the core of solenoid is ______ A/m. (A) 2.5×103 (B) 2×103 (C) 2.5×10−3 (D) 2×10−3
›Reveal solutionSolution
H=nI, independent of the core; H=1000×2=2×103 A/m. …
- GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQQ.A toroid wound with 100 turns/m of wire carries a current of 3A. The core of toroid is made of iron having relative magnetic permeability of mu_r = 5000 under given conditions. The magnetic field inside the iron is ___.(a) 0.15 T(b) 0.47 T(c) 1.5 x 10^-2 T(d) 1.88 T
›Reveal solutionSolution
The field in the toroid core is B = mu_0 mu_r n I; substituting gives about 1.88 T.
Field inside a toroid with a magnetic core: B = mu_0 mu_r n I, where n = 100 turns/m, I = 3 A, mu_r = 5000.
B = (4*pi x 10^-7)(5000)(100)(3) …
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