Magnetic Effects of Electric Current – Numerical Problems with Stepwise Solutions
CBSE • NCERT • Class 10
Physics — Chapter 12: Magnetic Effects of Electric Current
20 Numerical Problems with step-by-step solutions — topic-wise, NCERT-aligned for CBSE Class 10 board exam practice.
Includes
Field calculations, force (F = BIL), solenoid formulas, coil & loop problems, and simple induction/flux rate problems.
Content Bank — Key Formulas & Constants
- Magnetic field near long straight conductor: B = μ₀ I / (2π r)
- Field at centre of circular loop (single turn): B = μ₀ I / (2 r)
- Field inside long solenoid: B = μ₀ n I, where n = N / L (turns per metre)
- Force on straight conductor: F = B I L (if conductor perpendicular to B)
- Torque on rectangular coil (qualitative): τ ∝ B I A (A = area of loop)
- Permeability of free space: μ₀ = 4π × 10⁻⁷ T·m/A
- SI units: B in tesla (T), I in ampere (A), length in metre (m), force in newton (N).
Note: Class 10 problems emphasise proportional reasoning and unit conversions (cm→m, mT→T etc.).
Problem 1
Topic: Field of straight conductor
Q: A long straight wire carries a current of 5.0 A. Calculate the magnetic field at a point 4.0 cm away from the wire. (Use μ₀ = 4π×10⁻⁷ T·m/A)
1. Convert distance to metres: r = 4.0 cm = 0.04 m.
2. Use B = μ₀ I / (2π r).
3. Substitute values: B = (4π×10⁻⁷ × 5.0) / (2π × 0.04).
4. Simplify: B = (4π×10⁻⁷ × 5) / (2π × 0.04) = (20π×10⁻⁷) / (2π×0.04) = (10×10⁻⁷) / 0.04.
Result: B = (10×10⁻⁷) / 0.04 = 10×10⁻⁷ / 4×10⁻² = (10/4)×10^(−7+2) = 2.5×10⁻⁵ T = 25 μT.
Problem 2
Topic: Field vs distance (proportional)
Q: For the same wire as Problem 1, what will be the field at 8.0 cm? Explain relation.
1. Magnetic field for long wire B ∝ 1/r. Doubling r halves B.
2. From Problem 1, B(4cm) = 2.5×10⁻⁵ T. So at 8 cm B = 2.5×10⁻⁵ / 2 = 1.25×10⁻⁵ T.
Result: B = 1.25×10⁻⁵ T = 12.5 μT. (Shows inverse relation with r.)
Problem 3
Topic: Field at centre of circular loop
Q: A single circular loop of radius 10 cm carries a current of 2.0 A. Find the magnetic field at its centre. Use μ₀ = 4π×10⁻⁷ T·m/A.
1. Convert radius: r = 10 cm = 0.10 m.
2. Use B = μ₀ I / (2 r).
3. Substitute: B = (4π×10⁻⁷ × 2.0) / (2 × 0.10) = (8π×10⁻⁷) / 0.20.
4. Simplify: (8π×10⁻⁷)/(2×10⁻1) = 4π×10⁻⁶ / 10⁻1 = 4π×10⁻5 ≈ 4×3.14×10⁻5 = 12.56×10⁻5 T.
Result: B ≈ 1.256×10⁻⁴ T = 125.6 μT.
Problem 4
Topic: Multiple turns (coil)
Q: A coil has 50 turns, each of radius 5 cm. The coil carries 0.5 A. Estimate the field at centre assuming fields from turns add (use single loop formula multiplied by N).
1. r = 5 cm = 0.05 m. Single-turn B₁ = μ₀ I / (2 r) = (4π×10⁻⁷ × 0.5)/(2×0.05).
2. Compute single turn: numerator = 2π×10⁻7; denominator = 0.10 ⇒ B₁ = 2π×10⁻6 ≈ 6.283×10⁻6 T.
3. For N = 50 turns, B = N × B₁ = 50 × 6.283×10⁻6 = 3.1415×10⁻4 T.
Result: B ≈ 3.14×10⁻⁴ T = 314 μT.
Problem 5
Topic: Field inside solenoid
Q: A solenoid 0.40 m long has 800 turns and carries 1.5 A. Find the magnetic field inside the solenoid (assume ideal solenoid).
1. n = N/L = 800 / 0.40 = 2000 turns/m.
2. Use B = μ₀ n I = 4π×10⁻⁷ × 2000 × 1.5.
3. Compute: 4π×10⁻⁷ × 3000 = 12π×10⁻4 ≈ 12×3.14×10⁻4 = 37.68×10⁻4 = 3.768×10⁻3 T.
Result: B ≈ 3.77×10⁻³ T = 3.77 mT.
Problem 6
Topic: Field scaling with turns/current
Q: For the solenoid in Problem 5, if current is doubled, what is the new field? Explain proportionality.
1. B ∝ I for solenoid with fixed n. Original B ≈ 3.768×10⁻3 T at I = 1.5 A.
2. At 3.0 A (double), B_new = 2 × 3.768×10⁻3 = 7.536×10⁻3 T.
Result: B ≈ 7.54 mT. (Doubling current doubles B.)
Problem 7
Topic: Force on conductor (simple)
Q: A straight conductor of length 0.25 m carries a current of 4 A and is placed perpendicular to a uniform magnetic field of 0.2 T. Calculate the force on it.
1. Use F = B I L.
2. Substitute: F = 0.2 × 4 × 0.25 = 0.2 N.
Result: F = 0.20 N.
Problem 8
Topic: Force and angle
Q: Repeat Problem 7 if the conductor makes an angle of 30° with the magnetic field. (Use F = B I L sinθ)
1. Use F = B I L sinθ. Here θ = 30° and sin30° = 0.5.
2. F = 0.2 × 4 × 0.25 × 0.5 = 0.1 N.
Result: F = 0.10 N.
Problem 9
Topic: Force on moving charges (concept)
Q: Explain qualitatively why a current-carrying wire in a magnetic field experiences a force, while a stationary charge in a static magnetic field does not experience a force.
1. A current is a flow of charges (moving charges). A moving charge in a magnetic field experiences Lorentz force q(v×B). In a conductor, the collective effect appears as F = BIL.
2. A stationary charge (v = 0) has no magnetic force because q(v×B) = 0.
Result: Motion of charges is necessary for magnetic force — current in wire provides moving charges so wire experiences force in B.
Problem 10
Topic: Torque on rectangular loop (qualitative numeric)
Q: A rectangular coil of width 0.1 m and length 0.2 m carries 3 A in a magnetic field of 0.4 T. Calculate the force on one side of length 0.2 m (assume side is perpendicular to B) and estimate the turning effect qualitatively (torque ~ F × half-width).
1. Force on side of length L = 0.2 m: F = B I L = 0.4 × 3 × 0.2 = 0.24 N.
2. Approximate torque about center using lever arm ≈ half width = 0.1/2 = 0.05 m (if width = 0.1 m). τ ≈ F × 0.05 = 0.24 × 0.05 = 0.012 N·m.
Result: Force on side = 0.24 N; estimated torque ≈ 0.012 N·m (qualitative estimate for Class 10).
Problem 11
Topic: Field due to multiple conductors (superposition)
Q: Two long parallel wires separated by 0.10 m carry currents 6 A (wire A) and 4 A (wire B) in same direction. Find ratio of magnetic fields produced by A and B at a point midway between them.
1. Midpoint distance from each wire r = 0.10/2 = 0.05 m.
2. Use B ∝ I / r. For A: B_A ∝ 6 / 0.05 = 120. For B: B_B ∝ 4 / 0.05 = 80.
3. Ratio B_A : B_B = 120 : 80 = 3 : 2.
Result: Ratio = 3 : 2. (Fields depend on currents and same distance.)
Problem 12
Topic: Solenoid design (n and B)
Q: A solenoid is required to produce a field of 2.0×10⁻³ T when carrying 1.0 A. What turns per metre n are required? Use μ₀ = 4π×10⁻⁷.
1. Use B = μ₀ n I ⇒ n = B / (μ₀ I).
2. Substitute: n = 2.0×10⁻3 / (4π×10⁻7 × 1) = (2.0×10⁻3)/(4π×10⁻7).
3. Compute: 4π×10⁻7 ≈ 12.566×10⁻7 = 1.2566×10⁻6. So n ≈ 2.0×10⁻3 / 1.2566×10⁻6 ≈ 1592 turns/m.
Result: n ≈ 1.59×10³ turns per metre (about 1590 turns/m).
Problem 13
Topic: Induced emf — qualitative calculation
Q: A coil of 200 turns has its magnetic flux reduced uniformly from 4.0×10⁻³ Wb per turn to zero in 0.2 s. Estimate the average induced emf (magnitude).
1. Faraday's law (average emf magnitude) ≈ N × (ΔΦ / Δt).
2. ΔΦ per turn = 4.0×10⁻3 − 0 = 4.0×10⁻3 Wb. For N = 200, total flux change = 200 × 4.0×10⁻3 = 0.8 Wb.
3. Δt = 0.2 s ⇒ emf ≈ 0.8 / 0.2 = 4.0 V (average magnitude).
Result: Average induced emf ≈ 4.0 V.
Problem 14
Topic: Lenz's law — qualitative
Q: A magnet with north pole facing a coil is pushed towards it. State direction of induced current in coil (viewed from magnet side) according to Lenz's law and explain why.
1. As north pole approaches, flux through coil into coil increases.
2. Induced current will produce its own magnetic field to oppose increase — i.e., coil face will act like a north pole to repel approaching north pole.
3. Using right-hand rule for induced current (or Fleming's right-hand rule), the current will be anticlockwise when viewed from magnet side (classical convention — students should sketch to confirm).
Result: Induced current direction is such that coil's near face becomes north — it opposes approach (anticlockwise viewed from magnet side in this setup).
Problem 15
Topic: Generator emf (simple)
Q: A single loop rotates in a magnetic field so that the flux linked changes from +Φ to −Φ in 0.01 s. If the peak flux magnitude Φ = 1.5×10⁻4 Wb, estimate average emf magnitude induced during this interval.
1. ΔΦ = (−Φ) − (+Φ) = −2Φ ⇒ magnitude of change = 2Φ = 3.0×10⁻4 Wb.
2. Δt = 0.01 s ⇒ emf ≈ ΔΦ / Δt = 3.0×10⁻4 / 1.0×10⁻2 = 0.03 V.
Result: Average induced emf ≈ 3.0×10⁻2 V = 30 mV.
Problem 16
Topic: Force with multiple conductors
Q: A rectangular loop has two long sides each of length 0.3 m carrying current of 2 A in opposite directions. The loop lies in a uniform magnetic field B = 0.25 T perpendicular to the plane of the loop. Find net force on loop.
1. Forces on two long sides: F = B I L. Each side experiences equal magnitude F = 0.25 × 2 × 0.3 = 0.15 N but in opposite directions (because currents are opposite and field same).
2. These equal and opposite forces cancel, producing zero net force (but produce a torque if separated).
Result: Net force on loop = 0 N (forces on opposite sides cancel).
Problem 17
Topic: Field at centre—composite
Q: Two concentric circular loops carry currents 2 A (radius 4 cm) and 3 A (radius 6 cm) in same direction. Find net magnetic field at common centre (use single-loop approx B = μ₀ I / 2r).
1. Compute B1 = μ₀ I1 / (2 r1) with r1 = 0.04 m, I1 = 2 A.
2. B1 = (4π×10⁻7 × 2)/(2×0.04) = (8π×10⁻7)/(0.08) = 10π×10⁻6 ≈ 3.14×10⁻5 T.
3. Compute B2 for I2 = 3 A, r2 = 0.06 m: B2 = (4π×10⁻7 × 3)/(2×0.06) = (12π×10⁻7)/(0.12) = 10π×10⁻6 ≈ 3.14×10⁻5 T.
4. Net B = B1 + B2 (same direction) = 3.14×10⁻5 + 3.14×10⁻5 = 6.28×10⁻5 T.
Result: Net B ≈ 6.28×10⁻5 T = 62.8 μT.
Problem 18
Topic: Simple numerical on Lenz (energy)
Q: A coil has induced emf of 2 V when a magnet is moved. If the induced current in coil is 0.5 A, estimate the power dissipated in coil and comment on energy source.
1. Electrical power dissipated P = emf × current (approx) = 2 V × 0.5 A = 1.0 W.
2. This energy comes from mechanical work done to move magnet (you must do extra work against opposing induced magnetic force — Lenz's law), so mechanical energy converts to electrical (dissipated as heat).
Result: Power ≈ 1.0 W; energy supplied by work done on the system (mechanical input).
Problem 19
Topic: Field near pair of wires (net)
Q: Two long parallel wires 0.20 m apart carry equal currents 5 A in opposite directions. What is net magnetic field at midpoint between wires?
1. Midpoint distance from each wire r = 0.10 m. Field from each wire magnitude B = μ₀ I /(2π r) = 4π×10⁻7 ×5 /(2π×0.10) = (20π×10⁻7)/(0.2π) = (100×10⁻7) = 1.0×10⁻5 T.
2. Directions: Since currents are opposite, fields at midpoint from each wire add (they point same direction at midpoint). Thus net B = 2 × 1.0×10⁻5 = 2.0×10⁻5 T.
Result: Net B = 2.0×10⁻5 T = 20 μT.
Problem 20
Topic: Combined application & exam-style
Q: Design a short answer: A solenoid 0.5 m long has 1000 turns and carries 2 A. A straight wire of length 0.2 m carrying 3 A is placed at centre of solenoid perpendicular to axis inside the solenoid. Estimate magnetic field inside solenoid and force on wire (assume B uniform and perpendicular to wire). Use μ₀=4π×10⁻7.
1. n = 1000 / 0.5 = 2000 turns/m. B = μ₀ n I_solenoid = 4π×10⁻7 × 2000 × 2 = 16π×10⁻4 ≈ 5.0265×10⁻3 T.
2. Force on wire: F = B I_wire L = 5.0265×10⁻3 × 3 × 0.2 ≈ 5.0265×10⁻3 × 0.6 = 3.016×10⁻3 N.
Result: B ≈ 5.03×10⁻3 T (5.03 mT). Force ≈ 3.02×10⁻3 N (about 3.0 mN).