Skip to content
IV. Numerical Problems · Q16

Q.The solenoids S1S_1 and S2S_2 are wound on an iron-core of relative permeability 900. Their areas of cross-section and their lengths are the same and are 4 cm2^2 and 0.04 m respectively. If the number of turns in S1S_1 is 200 and that in S2S_2 is 800, calculate the mutual inductance between the solenoids. If the current in solenoid 1 is increased from 2A to 8A in 0.04 second, calculate the induced emf in solenoid 2.

Tamil Nadu DgeTextbookSubjectiveImportance★★★★★
51% · 67/132 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Step 1. Given: μr=900\mu_r=900, area =4 cm2=4×10−4 m2=4\ \text{cm}^2=4\times10^{-4}\ \text{m}^2, length l=0.04l=0.04 m (both solenoids share these); N1=200N_1=200 so n1=200/0.04=5000n_1=200/0.04=5000/m; N2=800N_2=800 so n2=800/0.04=20000n_2=800/0.04=20000/m.

Step 2. M=μ0μrn1n2Al=(4π×10−7)(900)(5000)(20000)(4×10−4)(0.04)M=\mu_0\mu_rn_1n_2Al = (4\pi\times10^{-7})(900)(5000)(20000)(4\times10^{-4})(0.04).

Step 3. Computing step by step: n1n2=5000×20000=108n_1n_2=5000\times20000=10^8; μ0μr=(4π×10−7)(900)≈1.1310×10−3\mu_0\mu_r=(4\pi\times10^{-7})(900)\approx1.1310\times10^{-3}; Al=4×10−4×0.04=1.6×10−5Al=4\times10^{-4}\times0.04=1.6\times10^{-5}; multiplying all: 1.1310×10−3×108×1.6×10−5≈1.8101.1310\times10^{-3}\times10^8\times1.6\times10^{-5} \approx 1.810 H.

Step 4. Current in solenoid 1 changes from 2A to 8A, so Δi1=6\Delta i_1=6 A over Δt=0.04\Delta t=0.04 s, giving Δi1/Δt=150\Delta i_1/\Delta t=150 A/s. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.