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Solenoid and Toroid

NEET > Physics > Magnetic Effects of Current and Magnetism > Magnetic Effect of Current > Solenoid and Toroid

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Overview content

Topic 5 of 8 • Chapter: Magnetic Effect of Current • Physics

Solenoid and Toroid – Complete Notes, Revision, Important Questions & Downloads

Solenoid and Toroid is built around the listed TOC subtopic Magnetic Field Due to Infinite Sheet Carrying Current, then extends the same Ampere-law symmetry to the long solenoid and the toroid. NEET tests this topic through direct formula questions such as B = mu0 j/2 for a current sheet, through inside-versus-end comparison in a long solenoid, and through toroid fields written as mu0NI/2pi r or mu0nI depending on the data supplied. A standard check is whether you can tell that the sheet field is uniform on both sides, the long-solenoid field is nearly uniform inside, and the toroid field is confined mainly within the core. These are short chapters on paper, but they punish any confusion between total turns N, turns per unit length n, and the correct region where the field exists.

⬇ Download Notes PDFView Important Questions →
Ampere GeometryStandard ResultsDirect Scoring
Expected QuestionsQ
1
question from current sheet, long solenoid, or toroid formula comparison
Time Required⏱
3 Hours
to lock the three standard geometries and the inside-outside field logic
Difficulty⚡
Medium
results are short, but wrong region selection or wrong use of n and N quickly flips the answer
NRI USA Curriculum GapUS
Moderate
international worksheets often stop at field lines qualitatively, while NEET expects fast formula recall and Ampere-loop reasoning
1Subtopics
28+Practice Questions
4Free Downloads
3 hrsPrep Time
⬇ Get Free Downloads

NEET Weightage & Exam Pattern

Magnetic Effect of Current
NEET YearQuestions from this TopicBarMarks
20241
 
1 Q
4
20231
 
1 Q
4
20221
 
1 Q
4
20212
 
2 Qs
8
20201
 
1 Q
4
Topic Weightage6 24
Infinite current sheet, long solenoid, and toroid are classic Ampere-law geometries because symmetry makes the field easy to read on a chosen loop.
NEET usually asks for one standard result, then adds a small trap about the correct region: above versus below sheet, inside versus end of solenoid, or inside versus outside toroid.

The most common algebraic slip is using total turns N where turns per unit length n is required, or forgetting that n = N/2pi r for a toroid of mean radius r.
📊
1.0
Avg Questions / Year
🎯
24
Total Marks (6 yrs)
📈
Direct
Pattern
⚠️
Medium
Difficulty

Preparation Strategy

1

Start with the Amperian Loop Shape For every formula in this topic, first picture the symmetry and the loop used in Ampere's law. This prevents blind memorisation and helps you remember why the sheet gives equal field on both sides while the toroid field depends on radius inside the core.

2

Separate the Three Geometries Cleanly Keep the infinite-sheet result, long-solenoid result, and toroid result in separate boxes in your notes. The trap is to treat them all as one formula family and forget which geometry produces uniform field, which has half-field at the end, and which depends on 1/r inside the ring.

3

Track n and N Without Guessing When the question gives turns per unit length, use n directly; when it gives total turns and mean radius, convert carefully before substituting. This is the fastest place to lose marks in toroid numericals.

4

Always Check the Region Before Substituting Ask whether the point is inside the solenoid, at one end, inside the toroid core, or outside it. The same formula is not valid in every region, and many NEET options are built exactly around that oversight.

Download Topic Notes

PDF · Cheat Sheet · MCQ Set · PYQ
📄
Full Topic Notes
Detailed notes on current sheet, long-solenoid field, end-field result, and toroid field inside the core.
PDF8 Pages
Download Notes
📝
Formula Sheet
One-page summary of mu0 j/2, mu0 nI, mu0 nI/2 at the end, and mu0NI/2pi r for a toroid.
PDF1 Page
Download Formulas
🎯
MCQ Practice
Practice set on region selection, field comparison, and turns-per-unit-length traps in solenoid and toroid questions.
PDF26 Questions
Download MCQs
⏳
Previous Year Questions
Selected PYQs on current-sheet symmetry, long-solenoid field, and toroid radius dependence.
PDF12 Questions
Download PYQs

Topic Coverage

2-Column Table
Column AColumn B
Magnetic Field Due to Infinite Sheet Carrying Current↗

Quick Revision

Concept → Trap → Example

1) Magnetic Field Due to Infinite Sheet Carrying Current

Ampere Family

For an infinite current sheet with linear current density j, symmetry and Ampere's law give B = mu0 j/2 on each side of the sheet. The same symmetry logic extends to a long solenoid with inside field mu0 nI and to a toroid with field mu0NI/2pi r inside the ring core.

  • The sheet field is uniform above and below the sheet and does not require a distance variable because symmetry fixes the magnitude at all nearby points on either side.
  • For a long solenoid, the field is nearly uniform inside, and at one end its magnitude becomes half the deep-inside value.
  • Trap: in a toroid, do not use the solenoid result blindly; first decide whether the point lies inside the core and whether the data are given as total turns N or turns per unit length n.
Example (NEET-style)If a sheet has j = 8 A/m, then the magnetic field on either side is mu0 x 8 divided by 2. For a long solenoid with n = 500 turns/m carrying 2 A, the inside field is mu0 x 500 x 2, while the field at one end is half of that value.

US Curriculum Gaps

Note for NRI/OCI students studying abroad.

Ampere-Law Symmetry Is Often Under-Drilled

Many school courses show field lines around solenoids qualitatively but do less explicit work on why symmetry makes current sheet, solenoid, and toroid into standard Ampere-law cases.

  • choosing a loop where B is constant
  • understanding why some line-integral segments contribute zero

Region-Based Formula Use

NEET expects students to decide whether a point is inside, outside, or at the end of a configuration before substituting a formula, especially for long solenoids and toroids.

  • half-field at the end of a long solenoid
  • field mainly confined to the toroid core

Concept IQ Check

Exam-style checks
1An infinite current sheet has linear current density j. The magnetic field magnitude on either side of the sheet is:Current sheet
mu0 j
mu0 j divided by 2
mu0 j divided by 4
zero
Applying Ampere's law to a rectangular loop that cuts through the sheet gives 2Bl = mu0 jl, so B = mu0 j/2. The field is uniform on both sides because symmetry makes the two contributing horizontal segments equivalent. Wrong options usually come from missing the factor of 2 created by those two equal path contributions.
2For a toroid of mean radius r with N turns carrying current I, the magnetic field inside the core is:Toroid
mu0 NI divided by 2pi r
mu0 I divided by 2r
mu0 nI divided by 2
independent of r
For a toroid, choosing a circular Amperian loop of radius r inside the core gives B(2pi r) = mu0 NI, so B = mu0 NI/2pi r. The solenoid-style result mu0 nI is also usable only after replacing n by N/2pi r. The other options mix the toroid with a circular loop or with the end field of a solenoid.

NEET Practice Questions

Click "Reveal Answer" after attempting
1A long solenoid has 400 turns per metre and carries a current of 3 A. The magnetic field well inside it is proportional to:
400 x 3
400/3
3/400
independent of both n and I
👁 Reveal Answer
400 x 3. For a long solenoid, the field inside is B = mu0 nI, so the magnitude is directly proportional to both turns per unit length and current. This question is checking whether you remember the proportionality correctly before bringing in mu0.
2If the field deep inside a long solenoid is B, what is the field near one end according to the standard result used here?
2B
B
B/2
0
👁 Reveal Answer
B/2. Near one end of a long solenoid, the textbook result gives half the deep-inside field. Students who answer B are usually carrying the inside formula into a boundary region without checking the geometry.
3A toroid carries current I. If the observation point moves to a larger radius within the core region, the magnetic field magnitude:
increases linearly with r
decreases as 1/r
remains exactly the same
becomes zero immediately
👁 Reveal Answer
Decreases as 1/r. Inside the toroid, B = mu0 NI/2pi r, so increasing the radius lowers the field inversely. The trap is to import the long-solenoid idea of perfectly uniform field into a geometry where the circular path length changes with radius.
4For an ideal infinite current sheet, the magnetic field directions on the two sides of the sheet are:
the same and parallel to the current
opposite and parallel to the sheet
both zero
radially outward from the sheet
👁 Reveal Answer
Opposite and parallel to the sheet. The current sheet produces uniform magnetic fields on both sides, but the directions reverse across the sheet according to the right-hand rule. Students who mark them as the same usually forget that the observation point has moved to the opposite side of the current distribution.
5A toroid has 500 turns, mean radius 0.1 m, and current 0.5 A. Which proportional expression matches the magnetic field inside the core?
500 x 0.5 divided by 0.1
0.1 divided by 500 x 0.5
500 divided by 0.5 x 0.1
independent of turns
👁 Reveal Answer
500 x 0.5 divided by 0.1. Inside a toroid, B is proportional to NI/r, so increasing turns or current increases the field while increasing mean radius reduces it. The proportional form therefore follows 500 x 0.5 divided by 0.1 before multiplying by the constant factor mu0/2pi.

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Frequently Asked Questions

Notes · Downloads · Revision · Important Questions
Why is the magnetic field of an infinite current sheet independent of distance in this model?
The symmetry of an ideal infinite sheet is the key reason. There is no preferred distance scale parallel to the sheet, so Ampere's law gives the same magnitude on either side once the enclosed current per unit length is fixed.
Why is a long solenoid treated as producing a nearly uniform field inside?
Deep inside a long solenoid, edge effects from the ends become negligible and the field lines are almost parallel to the axis. That symmetry makes the inside field nearly constant and allows the standard result B = mu0 nI.
What is the difference between a solenoid and a toroid?
A solenoid is a long helical cylindrical coil, while a toroid is essentially the same coil bent into a closed ring. The toroid has no free ends, so its field is mainly confined within the ring core region.
Why is the field near one end of a long solenoid half the deep-inside field?
At one end, the field contributions effectively come from only one side of the solenoid rather than from turns extending on both sides of the observation point. That reduces the standard deep-inside value by a factor of one-half.
What does n mean in solenoid and toroid formulas?
n means turns per unit length. In a solenoid it is N divided by the solenoid length, and in a toroid written with mean radius r it can be related through n = N/2pi r along the circular path.
Is the toroid field really zero everywhere outside?
In the ideal textbook model, the field outside the toroid is taken as negligible or zero because the net enclosed current for suitable outer Amperian loops is zero. Real coils can have small leakage fields, but NEET uses the idealised result.
Why is Ampere's law preferred over Biot-Savart's law here?
Both laws are valid, but the symmetry in current sheet, long solenoid, and toroid problems makes the line integral in Ampere's law much faster. Biot-Savart would be longer and less convenient for these geometries.
What is the most common trap in this topic?
Students often remember the right formula but use it in the wrong region or with the wrong form of turns data. Most incorrect answers come from confusing inside with end or outside, or from mixing N and n in toroid questions.
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Magnetic Field Due to Infinite Sheet Carrying Current

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Magnetic Field Due to Infinite Sheet Carrying Current

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