📌 1. Question Description
Ampere's Circuital Law
Ampere's circuital law relates the circulation of magnetic field around
a closed path to the total current enclosed by that path.
∮ B · dl = μ₀ Ienclosed
It is particularly useful for finding the magnetic field in systems having
high symmetry, such as:
- Long straight current-carrying conductor
- Long straight solenoid
- Toroidal coil
Most Important:
For a long ideal solenoid, the magnetic field inside is approximately
uniform while the field outside is nearly zero.
⚡ 2. Statement of Ampere's Circuital Law
The line integral of the magnetic field around any closed path is equal
to μ₀ times the net current enclosed by the path.
∮ B · dl = μ₀I
Here:
- B = magnetic field
- dl = small element of the closed path
- I = current enclosed by the path
- μ₀ = permeability of free space
📐 3. Mathematical Form
∮ B · d𝐥 = μ₀ Ienc
If the magnetic field is uniform and parallel to the path:
∮ B dl = B ∮dl
For a circular path of radius r:
B(2πr) = μ₀I
Therefore:
B = μ₀I / 2πr
🔬 4. Animated Solenoid & Toroid Practical
Current:
3.00 A
Turns / Length:
500
Radius:
20 cm
Magnetic Field:
3.77 × 10⁻³ T
MAGNETIC FIELD GENERATED ✔
🧲 5. Application: Magnetic Field Inside a Long Solenoid
Consider a long solenoid having N turns uniformly distributed over
length L and carrying current I.
Number of turns per unit length is:
n = N/L
Choose a rectangular Amperian loop of length l inside the solenoid.
Using Ampere's circuital law:
∮ B · dl = μ₀Ienc
Only the inside portion contributes significantly because the magnetic
field outside an ideal long solenoid is nearly zero.
For length l, the number of turns enclosed is:
nl
Therefore enclosed current is:
Ienc = nlI
Thus:
Bl = μ₀nlI
Therefore:
⭐ B = μ₀nI
Since n = N/L:
⭐ B = μ₀NI/L
🌐 6. Magnetic Field Outside a Long Solenoid
For an ideal infinitely long solenoid:
Outside magnetic field ≈ 0
The magnetic field lines are almost parallel and closely spaced inside
the solenoid, indicating a nearly uniform magnetic field.
⭕ 7. Application: Magnetic Field Inside a Toroid
A toroid is a solenoid bent into the form of a circular ring.
Suppose a toroid has N turns, carries current I and has mean radius r.
Choose a circular Amperian path of radius r inside the toroid.
By Ampere's circuital law:
∮ B · dl = μ₀Ienc
Since B is constant on the circular path:
B(2πr) = μ₀NI
Therefore:
⭐ B = μ₀NI / 2πr
📍 8. Magnetic Field in Different Regions of Toroid
Inside Central Hole
B = 0
No current is enclosed by the Amperian path.
Within Toroidal Core
B = μ₀NI/2πr
The Amperian path encloses all N turns.
Outside Toroid
B = 0
Net enclosed current is zero.
📊 9. Solenoid vs Toroid
| Feature |
Solenoid |
Toroid |
| Shape |
Cylindrical |
Circular ring |
| Field inside |
B = μ₀nI |
B = μ₀NI/2πr |
| Outside field |
Approximately zero |
Zero for ideal toroid |
| Field nature |
Nearly uniform inside |
Depends on radius |
📝 10. MCQ Practice
1. Ampere's circuital law is:
A. ∮B·dl = μ₀I
B. ∮B·dl = ε₀I
C. B = μ₀Q
D. B = Q/ε₀
✔ Answer: A
2. Magnetic field inside a long ideal solenoid is:
A. μ₀I/2πr
B. μ₀nI
C. μ₀I/r
D. Zero
✔ Answer: B
3. Number of turns per unit length of a solenoid is represented by:
A. Nl
B. N/L
C. L/N
D. NL
✔ Answer: B
4. Magnetic field outside an ideal infinitely long solenoid is:
A. Maximum
B. Infinite
C. Nearly zero
D. μ₀nI
✔ Answer: C
5. Magnetic field inside a toroid is:
A. μ₀NI/2πr
B. μ₀nI
C. μ₀IR
D. Zero
✔ Answer: A
6. A toroid is essentially a:
A. Straight wire
B. Circular loop
C. Solenoid bent into a circular ring
D. Capacitor
✔ Answer: C
7. The magnetic field inside a long solenoid depends upon:
A. Current
B. Number of turns per unit length
C. Both A and B
D. Radius only
✔ Answer: C
8. For a toroid, the magnetic field varies with:
A. r
B. 1/r
C. r²
D. 1/r²
✔ Answer: B
9. If current in a solenoid is doubled, its magnetic field becomes:
A. Half
B. Double
C. Four times
D. Unchanged
✔ Answer: B
10. SI unit of magnetic field is:
A. Tesla
B. Weber
C. Henry
D. Ampere
✔ Answer: A
🟣 11. Assertion–Reason — 5 Questions
Options:
A. Both A and R are true, and R is the correct explanation of A.
B. Both A and R are true, but R is not the correct explanation of A.
C. A is true, but R is false.
D. A is false, but R is true.
Assertion:
Magnetic field inside an ideal long solenoid is nearly uniform.
Reason:
The magnetic field lines inside a long solenoid are parallel and
closely spaced.
Answer: A
Assertion:
Magnetic field outside an ideal long solenoid is nearly zero.
Reason:
The magnetic field contributions outside the solenoid nearly cancel.
Answer: A
Assertion:
Magnetic field inside a toroid varies inversely with radius.
Reason:
For a toroid B = μ₀NI/2πr.
Answer: A
Assertion:
Ampere's law is especially useful for highly symmetric current
distributions.
Reason:
Symmetry allows the magnetic field to be taken outside the integral.
Answer: A
Assertion:
The magnetic field outside an ideal toroid is zero.
Reason:
An Amperian path outside the toroid encloses zero net current.
Answer: A
🟢 12. 2 Marks — 6 Questions
Q1
State Ampere's circuital law and write its mathematical expression.
Q2
What is meant by an Amperian loop?
Q3
Write the expression for magnetic field inside a long solenoid.
Q4
Define the number of turns per unit length of a solenoid.
Q5
Write the expression for magnetic field inside a toroid.
Q6
What is the magnetic field outside an ideal long solenoid?
🟡 13. 3 Marks — 6 Questions
Q1
State Ampere's circuital law and explain its physical meaning.
Q2
Derive the magnetic field inside a long solenoid using Ampere's law.
Q3
Explain why the magnetic field outside a long solenoid is nearly zero.
Q4
Derive the expression for magnetic field inside a toroid.
Q5
Explain the role of symmetry in applying Ampere's circuital law.
Q6
Differentiate between the magnetic field inside a solenoid and a toroid.
🟠 14. 4 Marks — 6 Questions
Q1
State Ampere's circuital law and derive the magnetic field inside a long
solenoid.
Q2
Using Ampere's law, derive the expression B = μ₀nI for a long solenoid.
Q3
Using a suitable Amperian loop, derive the magnetic field inside a toroid.
Q4
Explain the magnetic field in the three regions of a toroid.
Q5
A solenoid has 1000 turns per metre and carries 2 A current. Calculate
the magnetic field inside it.
Q6
A toroid has 500 turns, carries 2 A current and has mean radius 20 cm.
Calculate the magnetic field inside it.
🔴 15. 5 Marks — 6 Questions
Q1
State Ampere's circuital law and explain its application to a long
current-carrying solenoid.
Q2
Derive the magnetic field inside a long solenoid using a rectangular
Amperian loop.
Q3
Derive the magnetic field inside a toroid using Ampere's circuital law.
Q4
Explain why the field outside an ideal solenoid and outside an ideal
toroid is approximately zero.
Q5
Compare the magnetic fields of a solenoid and a toroid and derive their
respective expressions.
Q6
A solenoid has 2000 turns per metre and carries 3 A current. Calculate
the magnetic field inside it. Also explain the direction of the field.
🔵 16. 6 Marks — 6 Questions
Q1
State Ampere's circuital law and derive the magnetic field inside a long
solenoid with a neat labelled diagram.
Q2
Using Ampere's circuital law, derive the magnetic field inside a toroid
and discuss the field in all three regions.
Q3
Explain the complete application of Ampere's circuital law to a long
solenoid. Derive B = μ₀nI and discuss the field outside it.
Q4
Compare solenoid and toroid on the basis of construction, Amperian loop,
magnetic field and field distribution.
Q5
A toroid has 1000 turns, mean radius 25 cm and carries a current of
4 A. Calculate the magnetic field inside the toroid.
Q6
Explain Ampere's circuital law, its mathematical formulation and its
applications to both a long solenoid and a toroid.
🧮 17. Numerical Practice
Example 1 — Solenoid
A long solenoid has 1000 turns per metre and carries a current of 2 A.
Find the magnetic field inside it.
B = μ₀nI
B = 4π × 10⁻⁷ × 1000 × 2
B ≈ 2.51 × 10⁻³ T
✔ Answer: B ≈ 2.51 × 10⁻³ T
Example 2 — Toroid
A toroid has 500 turns, carries a current of 2 A and has mean radius
20 cm. Find the magnetic field inside it.
B = μ₀NI/2πr
r = 20 cm = 0.20 m
B =
(4π × 10⁻⁷ × 500 × 2)/(2π × 0.20)
B = 1.0 × 10⁻³ T
✔ Answer: B = 1.0 × 10⁻³ T
📚 18. Important Formula Sheet
Ampere's Circuital Law:
∮ B · dl = μ₀Ienc
Long Solenoid:
B = μ₀nI
Since n = N/L:
B = μ₀NI/L
Toroid:
B = μ₀NI/2πr
Outside Ideal Solenoid:
B ≈ 0
Outside Ideal Toroid:
B = 0
🚀 19. Quick Revision
🔹 Ampere's Law
∮B·dl = μ₀Ienc
🔹 Main Use
Useful for highly symmetric current distributions.
