Question

for a coil having self inductance 2mh.current flows at a rate of 10^3 Ampere/sec in it. find the emf induced in it​

Answer 1

Emf induced = -Ldi/dt

E = -2*10^-3*10^3

E = -2V

Answer 2

Answer:

For a coil having self inductance 2mh,current flows at a rate of 10³ Ampere/sec the emf induced in it​ is 2v.

Explanation:

A coil's self inductance is a property that opposes the change in current passing through it. The coil opposes the change in current by inducing an emf.

Relation gives the induced emf. 

  $emf = L\frac{di}{dt}$  

Given :        L = 2mh            

                  current flow rate($\frac{di}{dt}$) = 10³

so     $emf= L( 10^{3} )$    

        $emf = ( 2 \times 10^{-3} )(10^{3} )$

        $emf = 2 v$

Hence emf induced is 2v.