Emf induced in coil at t = 2 s when instantaneous flux linked is `phi = (5t^(3) - 100 t +300)` Wb:

Options

A. 300 V
B. 140 V
C. 40 V
D. `-40V`

Explanation

To find the electromotive force (emf) induced in the coil at \( t = 2 \) seconds, we will follow these steps:

### Step 1: Write down the expression for the instantaneous flux linked.
The instantaneous flux linked is given by:
\[ \phi(t) = 5t^3 - 100t + 300 \quad \text{(in Webers)} \]

### Step 2: Differentiate the flux with respect to time.
The induced emf (\( \mathcal{E} \)) in the coil is given by Faraday's law of electromagnetic induction, which states:
\[ \mathcal{E} = -\frac{d\phi}{dt} \]
Now, we differentiate \( \phi(t) \):
\[ \frac{d\phi}{dt} = \frac{d}{dt}(5t^3 - 100t + 300) \]
Using the power rule for differentiation, we get:
\[ \frac{d\phi}{dt} = 15t^2 - 100 \]

### Step 3: Substitute \( t = 2 \) seconds into the differentiated expression.
Now, we will substitute \( t = 2 \) seconds into the expression for \( \frac{d\phi}{dt} \):
\[ \frac{d\phi}{dt} \bigg|_{t=2} = 15(2)^2 - 100 \]
Calculating this gives:
\[ \frac{d\phi}{dt} \bigg|_{t=2} = 15(4) - 100 = 60 - 100 = -40 \]

### Step 4: Calculate the induced emf.
Now, we can find the induced emf using the formula:
\[ \mathcal{E} = -\frac{d\phi}{dt} \]
Substituting the value we found:
\[ \mathcal{E} = -(-40) = 40 \text{ V} \]

### Final Answer:
The emf induced in the coil at \( t = 2 \) seconds is \( 40 \) volts.

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