Question

A magnetic flux of 5 micro weber is linked with a coil when a current of 1mA flows through it . Calculate self inductance of the coil.

Answer 1

magnetic flux B=5mW=5x10^-6mw
current I =1mA=10^-3a
self inductance L=Q/I
L=5x10^-6/10^-3
L=5x10^-3Henry
L=5 mili Henry

Answer 2

Dear Student,

◆ Answer -

L = 5×10^-3 Wb/A

● Explanation -

# Given -

Φ = 5 uWb = 5×10^-6 Wb

i = 1 mA = 10^-3 A

# Solution -

Self inductance of the coil is calculated by formula -

L = Φ/i

L = 5×10^-6 / 10^-3

L = 5×10^-3 Wb/A

Hence, self inductance of the coil is 5×10^-3 Wb/A.

Thanks dear. Hope this helps you...