A long solenoid has 500 turns. When a current of 2 ~A is passed through it, the resulting magnetic flux linked with each turn of the solenoid is 4 10⁻³ ~Wb . The self-inductance of the solenoid is

Step-by-step solution

Key Idea : Inductance of a coil is numerically equal to the emf induced in the coil when the current in the coil changes at the rate of 1 As ⁻¹ . If I is the current flowing in the circuit, then flux linked with the circuit is observed to be proportional to I , ie, aligned & I & or =L I aligned where L is called the self-inductance or coefficient of self-inductance or simply inductance of the coil. Net flux through solenoid, =500 4 10⁻³=2 ~Wb or 2=L 2 [after putting values in Eq. or L=1 H