A closely wound coil of 100 turns and of crosssection 1 ~cm ^2 has coefficient of self inductance 1 mH . The magnetic induction at the centre of the core of a coil when a current of 2 A flows in it, will be (in Wb / m ^2 )

Step-by-step solution

Given that, N =100, ~A =1 ~cm ^2=1 10⁻⁴ ~m ^2 , L =1 mH =1 10⁻³ H and I =2 ~A Now, self-inductance of a coil is given by, L = ₀ ~N ^2 ~A 1 ...(i) Where ₀= permeability of free space =4 10⁻⁷ H / m and l is the length of the coil Also, magnetic induction B in the core, B = ₀ NI l From equation (i), aligned & l= ₀ ~N ^2 ~A ~L = (4 10⁻⁷ ) (100)^2 1 10⁻⁴ 1 10⁻³ & =4 10⁻⁷ 10^4 10⁻⁴ 10^3 & l=4 10⁻³ ~m aligned Substituting this in equation (ii), aligned B & = (4 10⁻⁷ ) 100 2 4 10⁻³ & =10⁻⁷ 200 10^3=0.2 ~Wb / m ^2 aligned