NEET2015PhysicsElectromagnetic InductionActual
The self inductance of a coil having 400 turns is 10 mH . The magnetic flux through the cross section of the coil corresponding to current 2 ~mA is
Options
- A2 10⁻⁵ ~Wb
- B2 10⁻³ ~Wb
- C3 10⁻⁵ ~Wb
- D8 10⁻³ ~Wb
Correct answer
A. 2 10⁻⁵ ~Wb
Step-by-step solution
Here, N=400, L=10 mH =10 10⁻³ H I=2 ~mA =2 10⁻³ ~A Total magnetic flux linked with the coil, aligned =N L I & =400 (10 10⁻³ ) 2 10⁻³ & =8 10⁻³ ~Wb aligned Magnetic flux through the cross-section of the coil = Magnetic flux linked with each turn = N = 8 10⁻³ 400 =2 10⁻⁵ ~Wb