Question

Question 6.9: A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?

Class 12 - Physics - Electromagnetic Induction Electromagnetic Induction Page-230

Answer 1

Given,
Mutual conductance, μ = 1.5 H
Initial current i₁= 0
Final current i₂ = 20 A
Time taken t = 0.5 seconds
Change in the current can be calculated as,di = 20 – 0
dI = 20 A
The mutual conductance of the circuit is calculated using the formula,
$e=\mu\frac{di}{dt}\\\mu=\frac{e}{\frac{di}{dt}}$
As we know,$e=\frac{d\phi}{dt}$

$\frac{d\phi}{dt}=\mu\frac{di}{dt}\\d\phi=\mu di$
= 1.5 H × 20A
= 30Wb

Hence, the change in flux linkage is 30Wb

Answer 2

Hey !!

Given :

Mutual inductance of a pair of coils, μ = 1.5 H

Initial current,              I₁ = 0 A

Final current,               I₂ = 20 A

Change in current will be:

              Δl = l₂ - l₁

             20 - 0 = 20 A

and we know,

                     ΔФ = MΔI,

where,

ΔФ is the change in the magnetic flux

                                       ΔФ = 1.5 × 30

                                             = 30 Wb

HENCE, CHANGE IN THE FLUX LINKAGE WILL BE 30 Wb

Good luck !!