Self and mutual inductance
- AB is a part of an electrical circuit (see figure). The potential difference "V_A - V_B", at the instant when current i = 2 A and is increasing at a rate of 1 amp / second is: Answer: 4. 10 volt Key idea: going from A to B, add the potential change across each element. The current flows from A to B and is increasin... 1
- The magnetic energy stored in an inductor of inductance 4 μH carrying a current of 2 A is : Answer: 1. 8 μ J Energy stored in an inductor: U=1/2LI² . U=1/2×(4×10⁻⁶)×(2)² = 8×10⁻⁶ J = 8 μ J. 1
- Two conducting circular loops of radii R_1 and R_2 are placed in the same plane with their centres coinciding. If R_1 >> R_2, the mutual inductance M between them will be directly proportional to : Answer: 4. R₂²/R₁ Key idea: the large loop R₁ produces a nearly uniform field at its centre, which passes through the small loop. B=μ₀I/2R₁... 1
- The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance Answer: 4. 13.89 H Energy stored in an inductor: U=1/2LI² . L=2U/I²=frac2×25×10⁻³(60×10⁻³)²=frac0.053.6×10⁻³≈13.89\ H 1
- A long solenoid of diameter 0.1 m has 2×10^4 turns per meter. At the centre of the solenoid, a coil of 100 turns and radius 0.01 m is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to 0A… Answer: 2. 32 μ C Key idea: charge q=ΔΦ/R (independent of the time taken). Field inside solenoid B=μ₀nI . Change: Δ B=4π×10⁻⁷×2×10⁴×4=0.032π... 1
- Figure shows a circuit that contains three identical resistors with resistance R = 9.0 Ω each, two identical inductors with inductance L = 2.0 mH each, and an ideal battery with emf ε = 18 V. The current 'i' through the battery just after the switch… Answer: Dropped in the official answer key: full marks to every candidate Key idea: just after the switch closes, an inductor acts as an ope... 1
- A long solenoid has 1000 turns. When a current of 4A flows through it, the magnetic flux linked with each turn of the solenoid is 4×10^-3 Wb. The self-inductance of the solenoid is : Answer: 4. 1 H Key idea: L=Nφ/I , where φ is the flux per turn. L=dfrac1000×4×10⁻³4=1 H. 1