A circular coil is placed perpendicular to the magnetic field. Calculate the emf induced in the coil at t = 2 s, if magnetic flux through coil varies as `phi=(10t^(2)+ 5t +5)` mWb.

Answer

To solve the problem of calculating the induced emf in a circular coil placed perpendicular to a magnetic field, we will follow these steps:

### Step-by-Step Solution

1. **Identify the given magnetic flux equation**:
The magnetic flux (Φ) through the coil is given by the equation:
\[ \Phi(t) = 10t^2 + 5t + 5 \quad \text{(in mWb)} \]

2. **Determine the expression for induced emf**:
The induced emf (ε) in the coil is related to the rate of change of magnetic flux through the coil. According to Faraday's law of electromagnetic induction, the induced emf is given by:
\[ \epsilon = -\frac{d\Phi}{dt} \]
For magnitude, we can write:
\[ |\epsilon| = \frac{d\Phi}{dt} \]

3. **Differentiate the magnetic flux with respect to time (t)**:
We need to calculate the derivative of the magnetic flux function:
\[ \frac{d\Phi}{dt} = \frac{d}{dt}(10t^2 + 5t + 5) \]
Using the power rule of differentiation:
- The derivative of \(10t^2\) is \(20t\).
- The derivative of \(5t\) is \(5\).
- The derivative of the constant \(5\) is \(0\).

Therefore, we have:
\[ \frac{d\Phi}{dt} = 20t + 5 \]

4. **Substitute \(t = 2\) seconds into the derivative**:
Now, we will substitute \(t = 2\) seconds into the expression for \(\frac{d\Phi}{dt}\):
\[ \frac{d\Phi}{dt} = 20(2) + 5 = 40 + 5 = 45 \quad \text{(in mWb/s)} \]

5. **Calculate the induced emf**:
Since we are interested in the magnitude of the induced emf:
\[ |\epsilon| = 45 \quad \text{(in volts)} \]

### Final Answer
The induced emf in the circular coil at \(t = 2\) seconds is:
\[ \epsilon = 45 \, \text{V} \]