✏️ Practice Section

Practice Zone — 3 Difficulty Levels

Easy (CBSE) → Moderate (NEET/JEE Main) → Advanced (JEE Advanced). Click options to reveal answers instantly. Track your score.

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📗 Set 1 — Easy (CBSE / NEET Basic)

5 questions • Foundational concepts

Easy

The SI unit of magnetic flux is:

Tesla (T)

Weber (Wb)

Henry (H)

Volt (V)

Weber (Wb) = T·m² = V·s. Tesla is unit of magnetic field B. Henry is inductance. Volt is EMF.

According to Faraday's law, the induced EMF in a coil is proportional to:

Magnetic flux through the coil

Rate of change of magnetic flux

Strength of the magnetic field

Number of field lines passing through

ε = −NdΦ/dt. EMF depends on RATE OF CHANGE (dΦ/dt), not the flux itself. Static flux = no EMF.

A bar magnet is held stationary inside a closed coil. The induced current is:

Maximum

Zero

Proportional to field strength

Proportional to coil resistance

Zero! Static magnet → constant flux → dΦ/dt = 0 → ε = 0 → I = 0. The most common EMI conceptual trap.

The induced charge through a circuit when flux changes from Φ₁ to Φ₂ with N turns and resistance R is:

N(Φ₂ − Φ₁) × Δt / R

N(Φ₂ − Φ₁) / R

R × N(Φ₂ − Φ₁)

N²(Φ₂ − Φ₁) / R

q = NΔΦ/R. Note: No time (Δt) in this formula. Induced charge is INDEPENDENT of time taken for flux change.

Self-inductance of a solenoid is doubled if (keeping all else constant):

Current is doubled

Length is doubled

Number of turns is increased by √2 times

Radius is doubled

L = μ₀N²A/l. L ∝ N². To double L, N must increase by √2 (since (√2)² = 2). L does not depend on current.

📙 Set 2 — Moderate (NEET / JEE Main)

5 questions • Application & calculation

Medium

A coil of 50 turns has flux changing from 0.04 Wb to 0.10 Wb in 0.06 s. The average induced EMF is:

3 V

10 V

50 V

100 V

ε = N × |ΔΦ/Δt| = 50 × (0.10−0.04)/0.06 = 50 × 0.06/0.06 = 50 × 1 = 50 V

A rod of length 0.8 m moves with velocity 5 m/s perpendicular to B = 2 T. The motional EMF is:

4 V

10 V

8 V

16 V

ε = BLv = 2 × 0.8 × 5 = 8 V

A solenoid (L = 5 mH) carries current 2 A. The energy stored in its magnetic field is:

20 mJ

5 mJ

10 mJ

2.5 mJ

U = ½LI² = ½ × 5×10⁻³ × 4 = 10 × 10⁻³ J = 10 mJ

A coil in uniform B rotates with ω = 100 rad/s. N = 50, A = 0.04 m², B = 0.5 T. Peak EMF is:

50 V

100 V

200 V

25 V

ε₀ = NBAω = 50 × 0.5 × 0.04 × 100 = 50 × 2 = 100 V

Q10

When a rod (L=1m, B=2T, v=3m/s) slides on rails with R=6Ω, the opposing force is:

6 N

2 N

12 N

4 N

F = B²L²v/R = 4×1×3/6 = 12/6 = 2 N. Or: ε=BLv=6V, I=6/6=1A, F=BIL=2×1×1=2N

📕 Set 3 — Advanced (JEE Advanced Level)

5 questions • Multi-concept, analytical

Hard JEE Adv

Q11

A rod of length L rotates with ω about one end in uniform B perpendicular to the plane. The EMF between the center and the free end is:

½BωL²

⅛BωL²

3BωL²/8

¼BωL²

EMF of full rod = ½BωL². EMF from pivot to center (length L/2): ε₁ = ½Bω(L/2)² = BωL²/8. EMF from center to end: ε₂ = ½BωL² − BωL²/8 = 4BωL²/8 − BωL²/8 = 3BωL²/8

Q12

Two inductors L₁ and L₂ are connected in series with mutual inductance M (opposing). The effective inductance is:

L₁ + L₂ + 2M

L₁ + L₂ − 2M

L₁ + L₂

L₁ + L₂ + M

Series with opposing mutual coupling: L_eff = L₁ + L₂ − 2M. Aiding: L₁ + L₂ + 2M. The ±2M term comes from the mutual EMF contributions when current flows through both coils.

Q13

A rod of mass m on rails (L, B ⊥ plane) is released from rest. If track resistance is R, the time constant for velocity build-up is:

L/R

mR/B²L²

mB²L²/R

mR/BL

From the differential equation: m(dv/dt) + (B²L²/R)v = mg. Time constant τ = m/(B²L²/R) = mR/B²L². Same form as RC circuit charging τ = RC.

Q14

In an LC circuit (L=4H, C=1μF), if maximum charge on capacitor is Q₀ = 2×10⁻³ C, the maximum current is:

1 A

0.5 A

2 A

0.25 A

I_max = Q₀/√(LC) = Q₀ × ω = (2×10⁻³)/√(4×10⁻⁶) = (2×10⁻³)/(2×10⁻³) = 1 A. Or: energy conservation ½LI²_max = Q₀²/2C → I_max = Q₀/√(LC)

Q15

Two rods (each mass m, resistance r) on resistanceless rails in field B (separation L). Rod A moves at v₀, B at rest. As t→∞, the total energy dissipated as heat is:

½mv₀²

¼mv₀²

mv₀²

⅛mv₀²

Both rods reach v_f = v₀/2 (momentum conservation). Initial KE = ½mv₀². Final KE = 2 × ½m(v₀/2)² = ¼mv₀². Heat = ½mv₀² − ¼mv₀² = ¼mv₀²

🎯 After Practice — Review Protocol

Score ≥ 80%: Ready for next difficulty level. Move to exam strategy.
Score 60–79%: Revisit specific topics in Core Concepts. Identify which questions you got wrong and which concept was involved.
Score < 60%: Return to Core Concepts and Formulas pages before attempting practice again. Don't rush difficulty levels.

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