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**Hint:**Use the formula of the emf given below and substitute the value of the induction and the equation of the current. Differentiate the current equation, and substitute the value of the time taken in it to find the emf at that corresponding time. The emf is directly proportional to time taken.

**Useful formula:**

The formula of the emf is given by

$ E = L\dfrac{{di}}{{dt}}$

Where $ E $ is the emf of the inductor, $L$ is the induction and $i$ is the current flowing through the inductor circuit.

**Complete step by step solution:**

It is given that the

The inductance of the inductor, \[L = 5\,mH = 5 \times {10^{ - 3}}\,H\]

The current through the inductor at time, $i = {t^2} + 4$

(i) Let us calculate the value of the emf through the inductor at $t = 1$ . First, write the formula of the emf,

$ E = L\dfrac{{di}}{{dt}}$

Substitute the known values in the above step,

$ E = 5 \times {10^{ - 3}} \times \dfrac{{d\left( {{t^2} + 4} \right)}}{{dt}}$

By performing differentiation in the above step,

$ E = 5 \times {10^{ - 3}} \times 2t$ -----(1)

Substituting the value of $t$ as $1\,{\text{second}}$ ,

$ E = 0.01\,V$

Let us calculate the emf for $t = 3\,s$ ,by substituting the value of $t$ as $3\,{\text{second}}$ ,

$ E = 5 \times {10^{ - 3}} \times 2 \times 3$

$ E = 30 \times {10^{ - 3}}$

$ E = 0.3\,V$

**Hence the emf of the inductor is obtained as $0.01\,V$ for $1\,\operatorname{s} $ time and $0.3\,V$ for the time period of $3\,s$ .**

(ii) The below graph shows the relation between the emf and the time taken for the current to flow. The emf is directly proportional to the current and thus the graph is straight line.

**Note:**The inductors can be functioned in two ways, first is to control signals and in the other is to store electrical energy. In the factor that helps to produce the emf, also generate the reactive fluxes that act against the produced emf.

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