Question

A solenoid has 2000 turns wound over a length of 0.30 m. Its area of cross-section is 1.2 × 10^–3 m²_.Around its central section a coil of 300 turns is wound. If an initial current of 2A in the solenoid is reversed in 0.25 sec, the emf induced in the coil is equal to –

• A. 6 × 10^–4 Volt
• B. 4.8 × 10^–2 Volt
• C. 6 × 10^–2 Volt
• D. 48 kV

Answer 1

Answer: (B)

4.8 × 10^–2 Volt

Answer 2

Magnetic field of solenoid, B₁ = $\frac{\mu_{0}N_{1}i_{1}}{\mathcal{l}}$

Magnet flux of coil, f₂ = N₂ B₁ A₂ = N₂ $\left( \frac{\mu_{0}N_{1}i_{1}}{\mathcal{l}} \right)$A₂

As f₂ = M i₁, so M = $\frac{\varphi_{2}}{i_{1}}$ = $\frac{\mu_{0}N_{1}N_{2}A_{2}}{\mathcal{l}}$

\ induced emf, |e| = M$\frac{di_{1}}{dt}$

or |e| = $\frac{\mu_{0}N_{1}N_{2}A_{2}}{\mathcal{l}}$ × $\frac{di_{1}}{dt}$

= $\frac{4\pi \times 10^{–7} \times 2000 \times 300 \times 1.2 \times 10^{- 3}}{0.30}$ × $\frac{4}{0.25}$

= 4.8 × 10^–2 Volt