A (current vs time) graph of the current passing through a solenoid is shown in Fig. For which time is the back electromotive force (u) a maximum? If the back emf t = 3 s is e, find the back emf at t = 7 s, 15 s and 40 s OA, AB and BC are straight line segments.

Explanation

Induced emf or back emf in a solenoid is given by
`|epsilon|=(LdI)/(dt)`
`epsilon` will be maximum when `(dI)/(dt)` is maximum.
In the graph shown, `(dI)/(dt)` (rate of change of current) is maximum for AB, i.e. for `5 s lt t lt 10 s`
(i) For `0 s lt t lt 5 s`, `(dI)/(dt)=((1-0)A)/((5-0)s)=(1)/(5) A // s`
Time t = 3 s will lie on the above slope. Back emf at t = 3 s is
`epsilon=-(LdI)/(dt)=-(L)/(5)` or L = - 5 e
(ii) For `5 s lt t lt 10 s`, `(dI)/(dt)=((-2-1)A)/((10-5)s)=-(3)/(5) A // s`
Time t = 7 s will lie on the above slope. Back emf at t = 7 s
`epsilon =-(LdI)/(dt)=-(-5e) xx ((-3)/(5))` (`:.` L = -5 e)
`implies epsilon= -3 e`
(iii) For `10 s lt t lt 30 s`
`(dI)/(dt)=([0-(-2)]A)/((30-10)s)=(1)/(10) A // s`
Time t = 15 s will lie on the above slope. Back emf at t = 15 s is
`epsilon=-(LdI)/(dt)= -(-5e) xx (1)/(10)`
`implies epsilon=(e)/(2)`
(iv) For `t gt 30 s`, `(dI)/(dt)=0`
Time t = 40 s will lie on the above slope. Back emf at t = 40 s is
`epsilon=-(LdI)/(dt)=0`