COMEDK2024Morning ShiftPhysicsElectromagnetic InductionActual
Around the central part of an air cored solenoid of length 20 ~cm and area of cross section 1.4 10⁻³ ~m ^2 and 3000 turns, another coil of 250 turns is closely wound. A current 2 ~A in the solenoid is reversed in 0.2 ~s , then the induced emf produced is
Options
- A1.32 10⁻¹ ~V
- B1.16 10⁻¹ ~V
- C4 10⁻¹ ~V
- D8 10⁻² ~V
Correct answer
A. 1.32 10⁻¹ ~V
Step-by-step solution
The magnetic field B inside a long solenoid is given by B = ₀ n I , where n = N l is the number of turns per unit length. Given: N₁ = 3000 , l = 0.2 m , I_ initial = 2 A , I_ final = -2 A , A = 1.4 10⁻³ m ^2 , N₂ = 250 , t = 0.2 s . The magnetic flux through the secondary coil is = B A = ₀ ( N₁ l ) I A . The change in magnetic flux is = _ final - _ initial = ₀ ( N₁ l ) A (I_ final - I_ initial ) . Substituting the values: = (4 10⁻⁷) ( 3000 0.2 ) (1.4 10⁻³) (-2 - 2) = (4 10⁻⁷) (15000) (1.4 10⁻³) (-4) = -1.055 10⁻⁴ W