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Magnetic Field Due to a Straight Wire

NEET > Physics > Magnetic Effects of Current and Magnetism > Magnetic Effect of Current > Magnetic Field Due to a Straight Wire

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Overview content

Topic 3 of 8 โ€ข Chapter: Magnetic Effect of Current โ€ข Physics

Magnetic Field Due to a Straight Wire โ€“ Complete Notes, Revision, Important Questions & Downloads

Magnetic Field Due to a Straight Wire packages General Formula and Special Cases into one usable family: finite wire, infinite wire, semi-infinite wire, axial point, and field inside or outside cylindrical conductors. NEET tests this topic through direct substitution in the standard formula, deciding the correct limiting case, and comparing field variation inside solid or hollow conductors. For example, once a long wire is treated as effectively infinite, the field varies directly with current and inversely with perpendicular distance, so doubling the distance halves the field.

โฌ‡ Download Notes PDFView Important Questions โ†’
Direct FormulaeHigh ROINEET Core
Expected QuestionsQ
1-2
questions from long-wire limits and cylindrical conductor cases
Time Requiredโฑ
2.5 Hours
to fix the special cases and the inside-outside behaviour
Difficultyโšก
Medium
standard results are simple, but students often choose the wrong geometric case
NRI USA Curriculum GapUS
Low
long-wire field is usually covered, but inside-cylinder current-distribution cases may be less emphasised
1Subtopics
32+Practice Questions
4Free Downloads
2.5 hrsPrep Time
โฌ‡ Get Free Downloads

NEET Weightage & Exam Pattern

Magnetic Effect of Current
NEET YearQuestions from this TopicBarMarks
20241
ย 
1 Q
4
20232
ย 
2 Qs
8
20221
ย 
1 Q
4
20211
ย 
1 Q
4
20202
ย 
2 Qs
8
Topic Weightage7ย 28
The infinite straight-wire result B = mu0 I divided by 2 pi r is one of the fastest one-step marks in this chapter.
NEET also checks whether the student can recognise when the semi-infinite or axial-position case should replace the infinite-wire formula.

Uniform current distribution inside solid and hollow cylinders is a common extension that tests Ampere-law reasoning instead of rote substitution.
๐Ÿ“Š
1.2
Avg Questions / Year
๐ŸŽฏ
28
Total Marks (6 yrs)
๐Ÿ“ˆ
Direct
Pattern
โš ๏ธ
Medium
Difficulty

Preparation Strategy

1

Identify the Wire Geometry First Before writing any formula, decide whether the conductor is finite, effectively infinite, semi-infinite, or a cylindrical current distribution. Most mistakes happen before algebra starts because the wrong case is chosen.

2

Track the Perpendicular Distance The field depends on the shortest perpendicular distance from the point to the wire, not simply on any labelled length in the figure. In NEET diagrams, mixing the actual perpendicular with some oblique segment is a standard trap.

3

Separate Inside and Outside Cylinder Results Outside a wire or cylinder, the field behaves like that of a long straight conductor carrying the total current. Inside a solid conductor, the enclosed current changes with radius; inside a hollow cylinder, the field can be zero in the hollow part.

Download Topic Notes

PDF ยท Cheat Sheet ยท MCQ Set ยท PYQ
๐Ÿ“„
Full Topic Notes
Notes covering finite, infinite, and semi-infinite wires plus solid and hollow cylindrical conductors.
PDF8 Pages
Download Notes
๐Ÿ“
Formula Sheet
Special-case sheet listing finite-wire angle form, infinite-wire result, and inside-cylinder formulae.
PDF1 Page
Download Formulas
๐ŸŽฏ
MCQ Practice
Practice set on distance dependence, field comparison at different radii, and conductor cross-section cases.
PDF32 Questions
Download MCQs
โณ
Previous Year Questions
Exam-style sheet focused on long straight wires, magnitude comparison, and field inside conductors.
PDF13 Questions
Download PYQs

Topic Coverage

2-Column Table
Column AColumn B
General Formula and Special Casesโ†—

Quick Revision

Concept โ†’ Trap โ†’ Example

1) General Formula and Special Cases

Wire Results

For a straight wire, the general field uses the angle form with the two end angles. It reduces to the infinite-wire result mu0 I divided by 2 pi r, the semi-infinite result mu0 I divided by 4 pi r, and zero at an axial point.

  • Use the full angle-based formula when the wire is finite and the observation point sees different end angles.
  • For a very long wire near its middle, both end angles become ninety degrees and the standard infinite-wire result follows immediately.
  • Trap: applying the infinite-wire formula to an end-point geometry where one of the limiting angles is actually zero.
Example (NEET-style)If a current of 10 A flows through a long straight wire and the point is 0.05 m away, the field is proportional to 10 divided by 0.05, so doubling the distance to 0.10 m cuts the field to half.

US Curriculum Gaps

Note for NRI/OCI students studying abroad.

Finite-Wire Angle Form

Many school courses emphasise only the infinite-wire limit, whereas NEET can still test the finite-wire angle expression and ask students to identify the correct limiting geometry.

  • reading end angles correctly
  • reducing the finite result to infinite or semi-infinite cases

Current Distribution Inside Conductors

Questions on solid, hollow, and thick hollow cylinders require enclosed-current reasoning that is often skipped in lighter introductory treatments.

  • uniform current density assumption
  • field zero in the hollow region

Concept IQ Check

Exam-style checks
1A point is at perpendicular distance r from a long straight wire carrying current I. If the distance is made 4r while current stays unchanged, what happens to the magnetic field?Distance scaling
It becomes four times.
It becomes one-fourth.
It becomes one-sixteenth.
It stays the same.
For an effectively infinite straight wire, the magnetic field varies inversely with the perpendicular distance r. Therefore increasing the distance from r to 4r reduces the field by the same factor 4. The one-sixteenth option comes from confusing this result with an inverse-square law. NEET uses this mistake often because students carry over the distance dependence from other field laws without checking the correct geometry.
2Inside the hollow region of a hollow cylindrical conductor carrying current uniformly in its material, the magnetic field is:Cylinder case
maximum
non-zero and constant
zero
inversely proportional to radius
Within the hollow region, an Amperian loop encloses no current, so the enclosed-current term is zero. That makes the magnetic field zero there. The field becomes non-zero only where current-carrying material is actually enclosed or outside the cylinder where the whole current is enclosed. This is a direct application of Ampere's Law and a common extension of straight-wire field questions in NEET practice.

NEET Practice Questions

Click "Reveal Answer" after attempting
1A long straight wire carries 20 A current. If a point is 0.1 m away from the wire, the magnetic field magnitude is proportional to:
20 divided by 0.1
20 times 0.1
0.1 divided by 20
1 divided by 20 times 0.1
๐Ÿ‘ Reveal Answer
20 divided by 0.1. For a long straight wire, the magnetic field magnitude is directly proportional to current and inversely proportional to perpendicular distance. Therefore the proportional part is I over r, which here is 20 divided by 0.1. The other options represent common confusions between direct and inverse dependence on distance.
2For a semi-infinite straight wire, the magnetic field at a point near the end is:
mu0 I divided by 2 pi r
mu0 I divided by 4 pi r
mu0 I divided by pi r
zero
๐Ÿ‘ Reveal Answer
Mu0 I divided by 4 pi r. In the semi-infinite case one end angle is ninety degrees and the other is zero, so only one sine term contributes in the general straight-wire expression. That leaves exactly half of the infinite-wire result. Students who write mu0 I divided by 2 pi r are treating the wire as infinite on both sides of the observation point, which is not the geometry described here.
3At a point lying on the axial line of a straight current-carrying conductor, the magnetic field is:
maximum
mu0 I divided by 2 pi r
zero
dependent only on current density
๐Ÿ‘ Reveal Answer
Zero. When the observation point lies on the axial position of the wire, the geometry makes the magnetic field contribution vanish because the relevant angular factor collapses. The textbook states this special case separately because it is easy to overlook when students try to apply the standard perpendicular-distance result everywhere.
4Inside a solid cylindrical conductor of radius R carrying uniform current I, the magnetic field at radius r inside the conductor is:
directly proportional to r
inversely proportional to r
independent of r
always zero
๐Ÿ‘ Reveal Answer
Directly proportional to r. For uniform current density, the enclosed current inside radius r scales as r squared, and Ampere's Law then gives B proportional to enclosed current divided by r. One factor of r remains, so the field rises linearly from the axis to the surface. This is why the inside-solid-cylinder graph does not have the same 1 over r behaviour as the outside region.

Physics Revision Checklist

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Use this section for quick chapter tracking before mocks, part tests, and final NEET revision.

Tip: Mark a chapter complete only after revising formulas, solving PYQs, and reviewing your error log for that chapter.

Frequently Asked Questions

Notes ยท Downloads ยท Revision ยท Important Questions
What is the safest formula to start with for a straight wire?
Start with the general finite-wire angle form if the ends matter in the geometry. Then reduce it to the infinite or semi-infinite limit only when the end angles justify that simplification.
Why does the field of a long straight wire vary as 1 over r and not 1 over r squared?
Because after integrating the Biot-Savart contributions of all current elements along the wire, the dependence changes from the current-element form to the long-wire result. The full geometry of the conductor matters.
How do I know a wire can be treated as infinite?
When the observation point is near the middle of a wire that is very long compared with the perpendicular distance, both end angles effectively become ninety degrees and the infinite-wire result is an excellent approximation.
Why is the field zero inside the hollow part of a hollow cylinder?
An Amperian loop drawn in the hollow region encloses no current. Since the enclosed current is zero, the magnetic field there is also zero.
What changes inside a solid cylindrical conductor?
Inside a solid conductor with uniform current density, the enclosed current grows with radius. That makes the magnetic field increase linearly with the distance from the axis until the surface is reached.
What happens to the field outside any cylindrical conductor carrying total current I?
Outside the conductor, the entire current is enclosed, so the field behaves like that of a long straight wire and varies as mu0 I divided by 2 pi r.
Why is the axial-position case separated in the textbook?
Because students often try to force the standard perpendicular-distance formula onto that geometry. On the axis of the conductor, the correct result is zero, so it must be recognised as a distinct special case.
What is the most common exam trap in this topic?
Using the infinite-wire result in every question without first checking whether the wire is finite, semi-infinite, on-axis, or part of a solid or hollow cylinder.
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General Formula and Special Cases

Subtopics

General Formula and Special Cases

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Magnetic Field Due to a Straight Wire > General Formula and Special Cases > Field Inside a Hollow Cylinder
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