CUET UG Physics Booster Test 3-Ampere\'s Circuital Law
📌 Answers are locked once submitted — results and explanations appear at the end.
QUESTION 1 OF 20
An Amperian loop encloses a current configuration such that ∮B⋅dl=12.56×10^(-6) T m. If μ_0=4π×10^(-7) T m A^(-1), what is the net enclosed current I_e? (Use π=3.14)
QUESTION 2 OF 20
Incorrect statement concerning the calculation of enclosed current in Ampere's law:
QUESTION 3 OF 20
Identify the correct statements regarding the open surface used in Ampere's law.
Statements:
1. It is conceptually akin to a soap film bounded by a wire loop.
2. Any shape of the surface will yield the same net current passing through it.
3. The magnetic field line integral must be evaluated over the entire open surface area.
4. The exact boundary of this surface is the Amperian loop itself.
QUESTION 4 OF 20
Identify the correct statements regarding the interaction of the boundary C and the open surface.
Statements:
1. The loop C must always be a perfect circle for Ampere's law to be mathematically valid.
2. The line integral is taken over the closed loop C coinciding with the boundary of the surface.
3. The open surface can be visually distorted without changing the boundary C.
4. The right-hand rule links the traversal sense of C to the positive normal of the surface.
QUESTION 5 OF 20
While Ampere's law holds for any loop, it may not always facilitate easy mathematical evaluation of the magnetic field unless:
QUESTION 6 OF 20
Match the component of Ampere's line integral with its condition.
| List I | List II |
|---|---|
| 1. ∫B⋅dl=BL | a. Magnetic field is perpendicular to every loop element |
| 2. ∫B⋅dl=0 | b. Magnetic field is tangential and constant along the loop |
| 3. B⋅dl=Bdl | c. Angle between Band dlis 0∘ |
| 4. B⋅dl=0 | d. Angle between Band dlis 90∘ |
QUESTION 7 OF 20
For an infinitely long straight wire, the field at every point on a circular loop of radius r centered on the wire is same in ________ and is directed ________ to the circle.
QUESTION 8 OF 20
The magnetic field of a long wire approaches ________ only when we come very close to the wire, demonstrating that it is ________ proportional to the radial distance.
QUESTION 9 OF 20
Identify the correct statements regarding cylindrical symmetry in Ampere's law.
Statements:
1. The magnetic field coordinates reduce from three spatial dimensions to just one.
2. Amperian loops are typically chosen as concentric circles.
3. The symmetry proves that the field at distance r is purely radial.
4. The field magnitude is independent of the polar angle around the axis.
QUESTION 10 OF 20
A magnetic field line loop forms a concentric circle of radius 10 cm around a long straight wire. If the line integral of B along this exact circular field line is 4π×10^(-6) T m, what is the uniform field magnitude at any point on this line?
QUESTION 11 OF 20
Which statement is incorrect regarding the right-hand rule used for Ampere's Law general applications?
QUESTION 12 OF 20
Identify the correct statements regarding the general right-hand rule used for the boundary sign convention in Ampere's Law.
Statements:
1. Fingers curl in the sense the boundary is traversed.
2. The thumb points in the direction of positive current.
3. If current flows opposite to the thumb, it is treated as negative in the integral.
4. The rule defines the orientation of the magnetic field independently of the loop.
QUESTION 13 OF 20
Identify the correct statements for an internal point r<a inside a solid wire of radius a carrying uniform current I.
Statements:
1. The Amperian loop chosen has radius r.
2. The total enclosed current is simply I.
3. The enclosed current is I(r^2/a^2).
4. The magnetic field is
B=μ_0Ir/2πa^2
QUESTION 14 OF 20
For r<a in a thick uniform wire, the plot of magnetic field magnitude B against distance r from the center of the wire is a ________ passing through the ________.
QUESTION 15 OF 20
A solenoid is mathematically and geometrically considered "long" when:
QUESTION 16 OF 20
Match the solenoid structural features with their physical effects.
| List I | List II |
|---|---|
| 1. Closely spaced neighbouring turns | a. Prevents electrical shorting between adjacent loops |
| 2. Enamelled wire coating | b. Allows each turn to be approximated as a circular current loop |
| 3. Long cylindrical winding | c. Produces a nearly uniform magnetic field inside |
| 4. Large number of turns | d. Strengthens the resultant magnetic field through superposition |
QUESTION 17 OF 20
A solenoid of length 0.5 m has 500 turns and carries a current of 5 A. Assuming it behaves as a long solenoid, what is the magnetic field deep inside?
μ_0=4π×10^(-7) T m A^(-1)
QUESTION 18 OF 20
For a rectangular Amperian loop abcd partially inside and outside a very long solenoid, the field along the outer edge cd is taken as ________, and along the transverse edges bc and ad, the field component is ________.
QUESTION 19 OF 20
Identify the correct statements comparing Ampere's and Gauss's macroscopic laws.
Statements:
1. Both relate boundary fields to interior sources.
2. Ampere's law exclusively uses a line integral on a one-dimensional boundary loop.
3. Both laws hold exclusively for rapidly changing fields.
4. Both are highly useful in continuous high-symmetry configurations.
QUESTION 20 OF 20
Identify the correct statements regarding the unification of physical concepts through Ampere's law.
Statements:
1. It provides a theoretical justification to Oersted's experiments.
2. It cannot be derived mathematically from the Biot–Savart law.
3. It proves that electric fields and magnetic fields are identical monopoles.
4. It simplifies the evaluation of fields in highly symmetric continuous current distributions.
Test Complete!
Answer Review
1 An Amperian loop encloses a current configuration such that ∮B⋅dl=12.56×10^(-6) T m. If μ_0=4π×10^(-7) T m A^(-1), what is the net enclosed current I_e? (Use π=3.14)
�� Apply Ampere's circuital law. �� Rearrange to find enclosed current. �� Substitute the given values.
Ampere's circuital law states ∮B⋅dl=μ_0I_e To determine the enclosed current, rearrange the equation: I_e=∮B⋅dl/μ_0 Substituting the given values, I_e=12.56×10^(-6)/4π×10^(-7) Using π=3.14, I_e=12.56×10^(-6)/12.56×10^(-7)I_e=10A Ampere's law directly relates the circulation of the magnetic field around a closed path to the total current enclosed by that path. The result depends only on the net enclosed current and not on the shape of the loop. Therefore, the enclosed current is 10 A.
- �� Option A → Corresponds to half the calculated current.
- �� Option C → Twice the correct value.
- �� Option D → Does not satisfy Ampere's law for the given integral value.
Substitution
- Application
- Use Ampere's circuital law and substitute the given numerical values.
- Final Logic
- I_e=12.56×10^(-6)/4π×10^(-7)=10A
"Integral ÷ μ₀ = Enclosed Current"
2 Incorrect statement concerning the calculation of enclosed current in Ampere's law:
�� Only enclosed currents contribute. �� Current signs are determined by the right-hand rule. �� Outside currents do not affect the integral.
Ampere's circuital law depends only on the net current enclosed by the chosen Amperian loop. Currents that pass outside the loop do not contribute to the enclosed current and therefore do not affect the value of the line integral. When multiple currents pass through the surface bounded by the loop, currents in opposite directions are assigned opposite signs and are algebraically added. The right-hand rule is used to determine which current direction is positive. For a thick wire carrying uniformly distributed current, the current enclosed within radius r is proportional to the area enclosed: I_(enc)=I(r^2/a^2) Thus Statements A, B and C are correct. The incorrect statement is that currents outside the loop contribute to the enclosed current.
- �� Option A → Oppositely directed currents must be algebraically subtracted.
- �� Option B → The right-hand rule determines the positive current direction.
- �� Option C → Correct expression for uniformly distributed current inside a thick conductor.
Concept Application
- Application
- Identify which currents are counted in Ampere's law.
- Final Logic
- Only currents passing through the surface bounded by the loop contribute.
"Outside Loop, Outside Calculation"
3 Identify the correct statements regarding the open surface used in Ampere's law.
Statements:
1. It is conceptually akin to a soap film bounded by a wire loop.
2. Any shape of the surface will yield the same net current passing through it.
3. The magnetic field line integral must be evaluated over the entire open surface area.
4. The exact boundary of this surface is the Amperian loop itself.
�� The surface is bounded by the loop. �� Different surfaces can share the same boundary. �� The integral is evaluated over the loop, not the surface.
The open surface associated with Ampere's law is often visualized as a soap film stretched across a wire loop. The chosen Amperian loop forms the boundary of this surface. The exact shape of the surface may be changed without altering the boundary loop, and the net enclosed current remains the same. Ampere's circuital law involves evaluating the magnetic field line integral around the boundary loop, not over the entire surface area. The surface is used only to determine which currents are enclosed. Therefore, Statements 1, 2 and 4 are correct, while Statement 3 is incorrect because the integration is performed along the closed loop rather than over the surface.
- �� Option A → Includes Statement 3, which is incorrect.
- �� Option B → Includes Statement 3, which is incorrect.
- �� Option D → Includes Statement 3 and excludes Statement 4.
NCERT Recall
- Application
- Recall the geometric interpretation of the surface and boundary in Ampere's law.
- Final Logic
- Surface determines enclosed current; loop determines the line integral.
"Soap Film Surface, Loop Integral"
4 Identify the correct statements regarding the interaction of the boundary C and the open surface.
Statements:
1. The loop C must always be a perfect circle for Ampere's law to be mathematically valid.
2. The line integral is taken over the closed loop C coinciding with the boundary of the surface.
3. The open surface can be visually distorted without changing the boundary C.
4. The right-hand rule links the traversal sense of C to the positive normal of the surface.
�� Ampere's law is valid for any closed loop. �� The boundary defines the line integral path. �� The right-hand rule determines orientation.
Ampere's circuital law is valid for any closed loop and does not require the loop to be circular. Circular loops are chosen only when symmetry simplifies calculations. The line integral is always evaluated along the closed boundary C of the chosen surface. The surface itself may be distorted into different shapes while maintaining the same boundary. The right-hand rule establishes the connection between the direction of traversal of the loop and the positive normal to the surface. This sign convention ensures consistency when determining the algebraic value of enclosed current. Therefore, Statements 2, 3 and 4 are correct, whereas Statement 1 is incorrect.
- �� Option B → Includes Statement 1, which is incorrect.
- �� Option C → Includes Statement 1, which is incorrect.
- �� Option D → Includes Statement 1, which is incorrect.
Logical Analysis
- Application
- Analyze the role of the boundary loop and the associated surface.
- Final Logic
- Ampere's law requires a closed loop, not necessarily a circular one.
"Any Closed Loop Works"
5 While Ampere's law holds for any loop, it may not always facilitate easy mathematical evaluation of the magnetic field unless:
�� Ampere's law is universally valid. �� Symmetry simplifies the integral. �� Circular, cylindrical and planar symmetries are most useful.
Ampere's circuital law is valid for any closed loop regardless of its shape. However, the law becomes especially useful when the current distribution possesses a high degree of symmetry. In systems such as infinitely long straight conductors, long solenoids and toroids, symmetry allows the magnetic field to be constant over selected portions of the Amperian loop. As a result, the line integral simplifies significantly and the magnetic field can be calculated directly. Without symmetry, Ampere's law remains valid but may not provide an easy method for evaluating the magnetic field. In such situations, the Biot–Savart law is often more convenient. Therefore, a high degree of symmetry is the key condition that makes Ampere's law practically useful.
- �� Option A → Loop size alone does not simplify the calculation.
- �� Option C → Time-dependent currents are not the condition for easy evaluation.
- �� Option D → Changing magnetic fields do not simplify Ampere's law calculations.
Concept Application
- Application
- Identify the physical condition that allows the line integral to be simplified.
- Final Logic
- High symmetry allows B to be constant or zero over parts of the loop, simplifying the integral.
"Symmetry Makes Ampere Easy"
6 Match the component of Ampere's line integral with its condition.
| List I | List II |
|---|---|
| 1. ∫B⋅dl=BL | a. Magnetic field is perpendicular to every loop element |
| 2. ∫B⋅dl=0 | b. Magnetic field is tangential and constant along the loop |
| 3. B⋅dl=Bdl | c. Angle between Band dlis 0∘ |
| 4. B⋅dl=0 | d. Angle between Band dlis 90∘ |
�� Tangential constant field gives BL. �� Perpendicular field gives zero contribution. �� Dot product depends on the angle between vectors.
Ampere's line integral depends on the relative orientation of the magnetic field B and the infinitesimal path element dl. When the magnetic field is tangential and constant throughout the loop, the line integral simplifies to BL. When the magnetic field is perpendicular to every element of the path, the dot product becomes zero because cos90^∘=0. Similarly, when the angle between B and dl is zero, the dot product becomes Bdl. These simplifications are extensively used in NCERT while applying Ampere's circuital law to highly symmetric systems. The correct matching therefore relates the integral expressions directly to the geometric conditions that produce them.
- �� Option A → Reverses the conditions for BL and zero integral.
- �� Option C → Interchanges the meanings of the dot-product conditions.
- �� Option D → Incorrectly matches tangential and perpendicular cases.
Concept Application
- Application
- Use the dot-product relation B⋅dl=Bdlcosθ.
- Final Logic
- Tangential Field → BL
- Perpendicular Field → 0
"Tangent Totals, Perpendicular Vanishes"
7 For an infinitely long straight wire, the field at every point on a circular loop of radius r centered on the wire is same in ________ and is directed ________ to the circle.
�� Cylindrical symmetry exists around the wire. �� Field magnitude is constant on a circle. �� Field direction is tangential to the circle.
An infinitely long straight current-carrying conductor exhibits cylindrical symmetry. Every point located at the same radial distance r from the wire experiences the same magnetic field magnitude. Therefore, all points on a circular path centered on the conductor have identical magnetic field magnitude. According to the right-hand thumb rule, the magnetic field lines form concentric circles around the wire. Consequently, the magnetic field direction at each point is tangential to the circular path. This property allows a circular Amperian loop to be chosen while applying Ampere's law, making the magnetic field constant throughout the loop and greatly simplifying calculations.
- �� Option A → The field is tangential, not normal.
- �� Option B → Magnitude remains constant, but direction changes from point to point.
- �� Option C → Neither condition is fully correct.
NCERT Recall
- Application
- Recall the symmetry of the magnetic field around a straight conductor.
- Final Logic
- Same Magnitude + Tangential Direction ⇒ Option D.
"Same Circle, Same Strength"
8 The magnetic field of a long wire approaches ________ only when we come very close to the wire, demonstrating that it is ________ proportional to the radial distance.
�� Field varies as 1/r. �� Smaller distance means larger field. �� At r→0, the field becomes very large.
The magnetic field due to a long straight conductor carrying current I is B=μ_0I/2πr This equation shows that the magnetic field is inversely proportional to the radial distance r. As the observation point moves closer to the conductor, the denominator becomes smaller and the magnetic field increases. In the ideal mathematical expression, the magnetic field tends toward infinity as r approaches zero. This behavior illustrates the inverse dependence of magnetic field on distance. Although a real conductor has finite radius and the expression is modified inside the conductor, the standard outside-field equation clearly demonstrates inverse proportionality.
- �� Option A → Field does not approach zero near the wire.
- �� Option B → First part is incorrect.
- �� Option C → Field is not directly proportional to distance.
Concept Application
- Application
- Analyze the formula B=μ_0I/2πr.
- Final Logic
- As r decreases, B increases; therefore B∝1/r.
"Near Wire, Field Climbs Higher"
9 Identify the correct statements regarding cylindrical symmetry in Ampere's law.
Statements:
1. The magnetic field coordinates reduce from three spatial dimensions to just one.
2. Amperian loops are typically chosen as concentric circles.
3. The symmetry proves that the field at distance r is purely radial.
4. The field magnitude is independent of the polar angle around the axis.
�� Cylindrical symmetry reduces coordinate dependence. �� Circular Amperian loops are chosen. �� Field direction is tangential, not radial.
For an infinitely long straight current-carrying conductor, cylindrical symmetry implies that the magnetic field magnitude depends only on the radial distance from the conductor. Thus, dependence on angular and axial coordinates disappears, reducing the problem to a single coordinate r. This symmetry allows concentric circular Amperian loops to be selected because the magnetic field remains constant on such loops. Furthermore, the field magnitude is independent of the polar angle around the axis. However, the field direction is not radial; it is tangential to the concentric circular field lines. Therefore, Statements 1, 2 and 4 are correct while Statement 3 is incorrect.
- �� Option B → Includes Statement 3, which is incorrect.
- �� Option C → Includes Statement 3.
- �� Option D → Includes Statement 3.
Logical Analysis
- Application
- Examine the consequences of cylindrical symmetry around a straight conductor.
- Final Logic
- Only radial distance determines magnitude; direction remains tangential.
"Only r Matters"
10 A magnetic field line loop forms a concentric circle of radius 10 cm around a long straight wire. If the line integral of B along this exact circular field line is 4π×10^(-6) T m, what is the uniform field magnitude at any point on this line?
�� Use ∮B dl=B(2πr). �� Radius is 10 cm = 0.1 m. �� Solve for B.
For a circular magnetic field line around a straight conductor, the magnetic field is tangential and constant throughout the circle. Therefore, ∮B⋅dl=B(2πr) Given ∮B⋅dl=4π×10^(-6) T m and r=10 cm=0.1 m Substituting, 4π×10^(-6)=B(2π×0.1)4π×10^(-6)=0.2πBB=4π×10^(-6)/0.2πB=20×10^(-6)B=2.0×10^(-5) T Hence, the magnetic field magnitude at every point on the circular field line is 2.0×10^(-5) T.
- �� Option A → Twice the calculated value.
- �� Option C → Ten times smaller than the correct value.
- �� Option D → Does not satisfy the given integral.
Substitution
- Application
- Apply the circular-loop relation ∮B dl=B(2πr).
- Final Logic
- B=4π×10^(-6)/2π(0.1)=2.0×10^(-5) T
"Integral ÷ Circumference = B"
11 Which statement is incorrect regarding the right-hand rule used for Ampere's Law general applications?
�� The thumb represents current direction in a straight wire. �� Fingers indicate magnetic field direction. �� Ampere's sign convention uses the same right-hand orientation.
For a long straight current-carrying conductor, the right-hand thumb rule states that the thumb points in the direction of current while the curled fingers indicate the direction of the magnetic field lines around the conductor. In Ampere's circuital law, the right-hand rule is also used to establish the sign convention between the traversal direction of the boundary loop and the positive direction of enclosed current. The fingers curl in the direction of traversal of the loop, while the thumb points along the positive current direction through the enclosed surface. Therefore, Statements A, B and D correctly describe applications of the right-hand rule. Statement C is incorrect because, for a straight wire, the thumb never points in the direction of the magnetic field; it points in the direction of current.
- �� Option A → Correctly describes the sign convention in Ampere's law.
- �� Option B → Correctly states the right-hand thumb rule for a straight conductor.
- �� Option D → Correctly relates loop traversal and positive current direction.
NCERT Recall
- Application
- Recall the standard right-hand thumb rule used for straight conductors.
- Final Logic
- Straight Wire → Thumb = Current, Fingers = Magnetic Field.
"Thumb Takes Current, Fingers Follow Field"
12 Identify the correct statements regarding the general right-hand rule used for the boundary sign convention in Ampere's Law.
Statements:
1. Fingers curl in the sense the boundary is traversed.
2. The thumb points in the direction of positive current.
3. If current flows opposite to the thumb, it is treated as negative in the integral.
4. The rule defines the orientation of the magnetic field independently of the loop.
�� Fingers determine loop traversal direction. �� Thumb defines positive current direction. �� Opposite current contributes negatively.
The right-hand rule used in Ampere's circuital law establishes the sign convention between the chosen traversal direction of the boundary loop and the enclosed current. If the fingers curl in the direction of traversal of the loop, the thumb points in the direction considered positive for the enclosed current. Any current flowing opposite to the thumb direction is assigned a negative sign in the enclosed current calculation. This convention ensures consistency between the line integral and the algebraic sign of the enclosed current. However, the rule does not independently define the magnetic field orientation without reference to the chosen loop and surface. Therefore, Statements 1, 2 and 3 are correct, while Statement 4 is incorrect.
- �� Option B → Includes Statement 4 and omits Statement 1.
- �� Option C → Includes Statement 4, which is incorrect.
- �� Option D → Includes Statement 4, which is incorrect.
Logical Analysis
- Application
- Analyze the role of fingers, thumb and current sign in Ampere's law.
- Final Logic
- Fingers → Loop Direction
- Thumb → Positive Current
- Opposite Current → Negative Sign
"Finger Loop, Thumb Current"
13 Identify the correct statements for an internal point r<a inside a solid wire of radius a carrying uniform current I.
Statements:
1. The Amperian loop chosen has radius r.
2. The total enclosed current is simply I.
3. The enclosed current is I(r^2/a^2).
4. The magnetic field is
B=μ_0Ir/2πa^2
�� A circular loop of radius r is selected. �� Only part of the current is enclosed. �� Field varies linearly with distance.
For a uniformly current-carrying solid conductor, Ampere's law is applied using a circular Amperian loop of radius r, where r<a. Since current density is uniform, the enclosed current is proportional to the enclosed area: I_(enc)=I(πr^2/πa^2)I_(enc)=I(r^2/a^2) Applying Ampere's circuital law, B(2πr)=μ_0I(r^2/a^2) which gives B=μ_0Ir/2πa^2 Thus, Statements 1, 3 and 4 are correct. Statement 2 is incorrect because the entire current I is enclosed only at the surface r=a, not for points inside the conductor.
- �� Option A → Statement 2 is incorrect.
- �� Option B → Includes Statement 2, which is false.
- �� Option D → Includes Statement 2 and omits Statement 4.
Concept Application
- Application
- Apply uniform current density and Ampere's law inside the conductor.
- Final Logic
- I_(enc)=I(r^2/a^2)B=μ_0Ir/2πa^2
"Inside Wire → Area First, Field Next"
14 For r<a in a thick uniform wire, the plot of magnetic field magnitude B against distance r from the center of the wire is a ________ passing through the ________.
�� Inside the conductor, B∝r. �� At the center, B=0. �� Linear dependence produces a straight line graph.
For a uniformly current-carrying thick conductor, the magnetic field inside the conductor is B=μ_0Ir/2πa^2 This equation shows that the magnetic field is directly proportional to the radial distance r. Therefore, when B is plotted against r, the graph is a straight line. At the center of the conductor (r=0), the magnetic field is zero because no current is enclosed within an infinitesimally small Amperian loop. As r increases, the enclosed current increases proportionally to the enclosed area, causing the magnetic field to increase linearly. Thus, the graph is a straight line passing through the origin.
- �� Option A → The relation is linear, not quadratic.
- �� Option C → Hyperbolic behavior occurs outside the conductor.
- �� Option D → Magnetic field is not constant inside the conductor.
Concept Application
- Application
- Use the relation B∝r for points inside the wire.
- Final Logic
- Linear relation ⇒ Straight Line Through Origin.
"Inside Wire, Draw a Line"
15 A solenoid is mathematically and geometrically considered "long" when:
�� A long solenoid has length much greater than radius. �� This produces a nearly uniform internal field. �� External field becomes negligible.
A solenoid is considered long when its length is much greater than its radius. Under this condition, edge effects become negligible and the magnetic field inside the solenoid becomes nearly uniform throughout most of its interior. At the same time, the magnetic field outside the solenoid becomes very weak and is often approximated as zero. This assumption is essential in deriving the standard expression B=μ_0nI for the magnetic field inside a long solenoid. The definition depends on geometry rather than current magnitude or turn spacing. Therefore, a solenoid is regarded as long when its length is significantly larger than its radius.
- �� Option A → Opposite of the required condition.
- �� Option B → Long solenoids have closely spaced turns.
- �� Option D → Solenoid length is independent of current magnitude.
NCERT Recall
- Application
- Recall the geometric condition used in the NCERT derivation of the magnetic field of a long solenoid.
- Final Logic
- Length ≫ Radius ⇒ Long Solenoid.
"Long Means Length Dominates"
16 Match the solenoid structural features with their physical effects.
| List I | List II |
|---|---|
| 1. Closely spaced neighbouring turns | a. Prevents electrical shorting between adjacent loops |
| 2. Enamelled wire coating | b. Allows each turn to be approximated as a circular current loop |
| 3. Long cylindrical winding | c. Produces a nearly uniform magnetic field inside |
| 4. Large number of turns | d. Strengthens the resultant magnetic field through superposition |
�� Closely spaced turns behave like continuous loops. �� Enamel insulation prevents short-circuiting. �� Long solenoids produce nearly uniform fields.
A solenoid consists of a long insulated wire wound in the form of a closely packed helix. Closely spaced turns allow each turn to be treated approximately as a circular current loop whose magnetic field contributes to the overall field. The enamel coating on the wire prevents electrical contact between neighbouring turns and avoids short-circuiting. When the solenoid length is large compared to its radius, the magnetic field inside becomes nearly uniform. A large number of turns causes the magnetic fields of individual loops to add through superposition, producing a stronger resultant field. These structural features are essential for the operation of practical solenoids and are emphasized in NCERT while discussing magnetic field generation by current-carrying coils.
- �� Option A → Incorrectly exchanges the functions of closely spaced turns and enamel coating.
- �� Option B → Incorrectly interchanges the effects of length and number of turns.
- �� Option C → Incorrectly matches the role of long cylindrical winding.
NCERT Recall
- Application
- Recall the construction features of a solenoid and their physical significance.
- Final Logic
- Turns → Circular Loop Approximation
- Enamel → Insulation
- Long Length → Uniform Field
- More Turns → Stronger Field
"Close, Coat, Long, Many"
17 A solenoid of length 0.5 m has 500 turns and carries a current of 5 A. Assuming it behaves as a long solenoid, what is the magnetic field deep inside?
μ_0=4π×10^(-7) T m A^(-1)
�� Use the long solenoid field formula. �� Calculate turns per unit length. �� Substitute numerical values.
For a long solenoid, the magnetic field inside is given by B=μ_0nI where n=N/L is the number of turns per unit length. Given: N=500,L=0.5 mn=500/0.5=1000 m^(-1) Substituting into the field equation, B=(4π×10^(-7))(1000)(5)B=20π×10^(-4)B=6.28×10^(-3) T Thus, the magnetic field deep inside the solenoid is 6.28×10^(-3) T. This result follows directly from Ampere's circuital law applied to a long solenoid.
- �� Option B → Half the correct value.
- �� Option C → Incorrect substitution of turns per unit length.
- �� Option D → Much smaller than the calculated field.
Substitution
- Application
- Use B=μ_0nI with n=N/L.
- Final Logic
- n=1000 m^(-1)B=6.28×10^(-3) T
"Field Inside = μ₀ × Turns Density × Current"
18 For a rectangular Amperian loop abcd partially inside and outside a very long solenoid, the field along the outer edge cd is taken as ________, and along the transverse edges bc and ad, the field component is ________.
�� External field of a long solenoid is negligible. �� Transverse segments are perpendicular to the field. �� Their contribution to the integral vanishes.
To derive the magnetic field inside a long solenoid using Ampere's law, a rectangular Amperian loop is chosen. One side lies inside the solenoid parallel to the magnetic field, while another side lies outside where the magnetic field is approximately zero. Therefore, the contribution from the outer edge cd is taken as zero. The transverse edges bc and ad are perpendicular to the magnetic field direction. Since B⋅dl=Bdlcos90^∘ their contributions are also zero. As a result, only the segment inside the solenoid contributes significantly to the line integral. This simplification leads directly to the expression B=μ_0nI.
- �� Option A → Neither contribution is maximum.
- �� Option B → Outer edge is not uniform; it is approximately zero.
- �� Option D → Transverse segments contribute zero, not maximum.
Concept Application
- Application
- Examine the contribution of each side of the Amperian rectangle.
- Final Logic
- Outer Edge → Zero
- Transverse Edges → Zero
"Outside Fades, Sideways Cancels"
19 Identify the correct statements comparing Ampere's and Gauss's macroscopic laws.
Statements:
1. Both relate boundary fields to interior sources.
2. Ampere's law exclusively uses a line integral on a one-dimensional boundary loop.
3. Both laws hold exclusively for rapidly changing fields.
4. Both are highly useful in continuous high-symmetry configurations.
�� Both connect boundary quantities with enclosed sources. �� Ampere's law uses a line integral. �� Symmetry makes both laws powerful.
Ampere's circuital law relates the circulation of the magnetic field around a closed boundary loop to the current enclosed by that loop. Gauss's law relates the electric flux through a closed surface to the enclosed electric charge. Thus, both laws connect a quantity defined on a boundary to a source located within the enclosed region. Ampere's law specifically employs a line integral along a one-dimensional closed path. Both laws are particularly useful for highly symmetric systems such as infinite wires, solenoids, spherical charge distributions and cylindrical charge distributions. Statement 3 is incorrect because the standard NCERT forms are not restricted exclusively to rapidly varying fields.
- �� Option A → Includes Statement 3, which is incorrect.
- �� Option B → Includes Statement 3.
- �� Option C → Includes Statement 3.
Logical Analysis
- Application
- Compare the mathematical structures of Ampere's and Gauss's laws.
- Final Logic
- Boundary Quantity ↔ Interior Source
- Symmetry Makes Both Powerful
"Boundary Reads the Source"
20 Identify the correct statements regarding the unification of physical concepts through Ampere's law.
Statements:
1. It provides a theoretical justification to Oersted's experiments.
2. It cannot be derived mathematically from the Biot–Savart law.
3. It proves that electric fields and magnetic fields are identical monopoles.
4. It simplifies the evaluation of fields in highly symmetric continuous current distributions.
�� Ampere's law explains magnetic effects of current. �� It supports Oersted's observations. �� It is powerful for symmetric current distributions.
Oersted's experiment demonstrated that an electric current produces a magnetic field. Ampere's circuital law provides a quantitative and theoretical framework for understanding this observation by relating magnetic field circulation to enclosed current. The law is especially useful for continuous current distributions possessing high symmetry, such as infinitely long conductors and long solenoids, where direct magnetic field evaluation becomes straightforward. Statement 2 is incorrect because Ampere's law can be derived from the Biot–Savart law under steady-current conditions. Statement 3 is also incorrect because electric and magnetic fields are distinct physical entities, and magnetic monopoles have not been experimentally observed. Therefore, only Statements 1 and 4 are correct.
- �� Option B → Statement 2 is incorrect.
- �� Option C → Statements 2 and 3 are incorrect.
- �� Option D → Statement 3 is incorrect.
Concept Application
- Application
- Analyze the physical significance and limitations of Ampere's law.
- Final Logic
- Ampere's Law Explains Oersted's Result and Simplifies Symmetric Field Calculations.
"Oersted Observed, Ampere Explained"
