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Class 12  »  Physics  »  Moving Charges and Magnetism

Amperes Circuital Law - UNSOLVED PRACTICE SET

Class 12

Chapter: Moving Charges and Magnetism | Topic: Amperes Circuital Law

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Class 12

AMPERES CIRCUITAL LAW - UNSOLVED PRACTICE SET

Topic: Amperes Circuital Law

Time: 40 mins | Marks: 30 | Difficulty: Medium

Multiple Choice Questions

Q1. Ampere's Circuital Law states that:

  1. ∮B · dl = μ₀I_enclosed
  2. ∮B · dl = I_enclosed/μ₀
  3. ∮B × dl = μ₀I_enclosed
  4.  ∮B · dl = 0

Q2. Ampere's Circuital Law is most useful for calculating magnetic fields when:

  1. The current distribution is asymmetric
  2. The current distribution has high symmetry
  3. There are no currents
  4. The magnetic field is zero

Q3. The line integral of magnetic field around a closed loop is zero when:

  1. No current passes through the loop
  2. The net current enclosed is zero
  3. Both (a) and (b)
  4. The loop is circular

Q4. For a long straight wire carrying current I, the Amperian loop is chosen as:

  1. A square around the wire
  2. A circle centered on the wire
  3. A rectangle not enclosing the wire
  4. Any shape, but the field is hard to calculate

Q5. Ampere's Circuital Law is analogous to:

  1. Coulomb's Law
  2. Gauss's Law in electrostatics
  3. Ohm's Law
  4. Biot-Savart Law

Q6. The magnetic field inside a long straight wire of radius R carrying uniform current I at distance r from the center (r < R) is:

  1. μ₀I/2πr
  2. μ₀Ir/2πR²
  3. μ₀I/2πR
  4. Zero

Short Answer Questions

Q7. State Ampere's Circuital Law in mathematical form. What does each term represent?

Q8. Why is Ampere's Circuital Law easier to use than Biot-Savart Law for a long straight wire? Explain.

Q9. A long straight wire carries current 20 A. Using Ampere's Circuital Law, calculate the magnetic field at distance 8 cm from the wire.

[Given: μ₀ = 4π × 10⁻⁷ T m/A]

Q10. Explain why the magnetic field inside a hollow cylindrical conductor carrying current is zero.

Q11. An Amperian loop encloses two wires carrying currents 5 A and 3 A in opposite directions. What is the value of ∮B · dl around this loop?

Q12. Compare Ampere's Circuital Law with Gauss's Law in electrostatics. List two similarities and two differences.

Long Answer Questions

Q13. State and prove Ampere's Circuital Law. Use it to derive the expression for the magnetic field due to a long straight current-carrying wire at perpendicular distance r from it. Draw the Amperian loop and explain each step.

Q14. Using Ampere's Circuital Law, derive the expression for the magnetic field inside a long straight wire of radius R carrying uniform current density. Show that the field increases linearly with distance inside the wire and decreases inversely outside.

Q15. A coaxial cable has an inner conductor of radius a = 2 mm carrying current I = 10 A upward, and an outer cylindrical shell of radius b = 5 mm carrying the same current downward (uniformly distributed).

(a) Calculate the magnetic field at r = 1 mm (inside the inner conductor).  

(b) Calculate the field at r = 3 mm (between the conductors).  

(c) Calculate the field at r = 6 mm (outside the cable).  

(d) Sketch a graph of B versus r from r = 0 to r = 8 mm.

[Given: μ₀ = 4π × 10⁻⁷ T m/A]

Numerical / Application-Based Problems

Q16. In a school physics lab, a student verifies Ampere's Circuital Law using a long straight wire carrying I = 15 A.

(a) Calculate ∮B · dl for a circular Amperian loop of radius 5 cm centered on the wire.  

(b) Verify that this equals μ₀I_enclosed.  

(c) The student now uses a square loop of side 10 cm centered on the wire. Calculate ∮B · dl for this loop.  

(d) For the square loop, calculate B at each side and show that the line integral still gives μ₀I.  

(e) A student asks: "If I choose a loop that doesn't enclose the wire, what is ∮B · dl?" Explain clearly.

[Given: μ₀ = 4π × 10⁻⁷ T m/A]

Q17. A solid cylindrical conductor of radius R = 1 cm carries current I = 50 A distributed uniformly across its cross-section.

(a) Using Ampere's Circuital Law, derive B(r) for r < R and r > R.  

(b) Calculate B at r = 0.5 cm, 1 cm, and 2 cm.  

(c) Plot a graph of B(r) versus r from 0 to 3 cm.  

(d) Calculate the current density in the conductor.  

(e) A hollow cylindrical conductor with inner radius 0.5 cm and outer radius 1 cm carries the same total current. How does the field at r = 0.75 cm differ from the solid conductor case? Explain.

[Given: μ₀ = 4π × 10⁻⁷ T m/A]

Q18. A toroid has N = 500 turns, mean radius R = 20 cm, and cross-sectional radius r = 2 cm. It carries current I = 3 A.

(a) Using Ampere's Circuital Law, derive the expression for the magnetic field inside the toroid.  

(b) Calculate the magnetic field at the mean radius.  

(c) Calculate the field at the inner edge and outer edge of the toroid.  

(d) Show that the field outside the toroid (in the central hole and outside the windings) is zero.  

(e) Compare the toroid's field with that of a solenoid bent into a circle. Why is the toroid's field more confined?

[Given: μ₀ = 4π × 10⁻⁷ T m/A]


Total: 30 Marks | Time: 40 mins

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