Question
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
2 × 10-5 Wb
2 × 10-3 Wb
3 × 10-5 Wb
8 × 10-3 Wb
Solution
A.
2 × 10-5 Wb
Number of turns N = 400
L = 10 mH = 10 × 10-3 H
I = 2 mA = 2 × 10-3 A
Total magnetic flux linked with the coil,
= N L I
= 400 × ( 10 × 10-3 ) × 2 × 10-3
= 8 × 10-3 Wb
Magnetic flux through the cross-section of the coil = Magnetic flux linked with each turn
=
= 2 × 10-5 Wb
