Let the current be 2A at t=τ
Then 2=2sinτ2⇒τ=π/2
when the instantaneous current is i, the
self-induced emf is Ldtdi. If dq amount of
charge is displaced in time dt, then elementary work done
=(Ldtdi)dq=Ldtdiidt=Lidi
∴W=∫0tLidi =∫0tL2sint2d(2sint2)
⇒W=∫0t8Lsint2cost2tdt=4L∫0tsin2t2dt
Let θ=2t2⇒dθ=4tdt
∴ the integral
=4L∫sinθdt/4L(−cosθ)=−Lcos2t2
∴W=−[cos2t2]0π/2=2L=4J