The flux of magnetic field through a closed conducting loop of resistance `0.4 Omega` changes with time according to the equation `Phi =0.20 t^(2)+0.40t+0.60` where `t` is time in seconds.Find (i)the induced `emf` at `t=2s`.(ii)the average induced `emf` in `t=0` to `t=5 s`.(iii)charge passed through the loop in `t=0` to `t=5s` (iv)average current.In time interval `t=0` to `t=5 s` (v) heat produced in `t=0` to `t=5s`

Explanation

To solve the given problem step by step, we will follow the instructions provided in the video transcript.

### Given:
- Resistance \( R = 0.4 \, \Omega \)
- Magnetic flux \( \Phi(t) = 0.20 t^2 + 0.40 t + 0.60 \)

### (i) Find the induced emf at \( t = 2 \, s \)

1. **Calculate the induced emf using Faraday's law**:
\[ \text{Induced emf} (E) = -\frac{d\Phi}{dt} \]
Differentiate \( \Phi(t) \):
\[ \frac{d\Phi}{dt} = \frac{d}{dt}(0.20 t^2 + 0.40 t + 0.60) = 0.40 t + 0.40 \]
Therefore,
\[ E = - (0.40 t + 0.40) \]

2. **Substitute \( t = 2 \, s \)**:
\[ E = - (0.40 \times 2 + 0.40) = - (0.80 + 0.40) = -1.20 \, \text{V} \]

### (ii) Find the average induced emf from \( t = 0 \) to \( t = 5 \, s \)

1. **Calculate \( \Phi(5) \) and \( \Phi(0) \)**:
\[ \Phi(5) = 0.20 \times 5^2 + 0.40 \times 5 + 0.60 = 0.20 \times 25 + 2 + 0.60 = 5 + 2 + 0.60 = 7.60 \]
\[ \Phi(0) = 0.20 \times 0^2 + 0.40 \times 0 + 0.60 = 0.60 \]

2. **Calculate the change in flux**:
\[ \Delta \Phi = \Phi(5) - \Phi(0) = 7.60 - 0.60 = 7.00 \]

3. **Calculate the average emf**:
\[ \text{Average emf} = \frac{\Delta \Phi}{\Delta t} = \frac{7.00}{5 - 0} = \frac{7.00}{5} = 1.40 \, \text{V} \]

### (iii) Find the charge passed through the loop from \( t = 0 \) to \( t = 5 \, s \)

1. **Using the formula for charge**:
\[ Q = \frac{\Delta \Phi}{R} \]
Substitute \( \Delta \Phi \) and \( R \):
\[ Q = \frac{7.00}{0.4} = 17.5 \, \text{C} \]

### (iv) Find the average current in the time interval \( t = 0 \) to \( t = 5 \, s \)

1. **Calculate average current**:
\[ I_{\text{avg}} = \frac{Q}{\Delta t} = \frac{17.5}{5} = 3.5 \, \text{A} \]

### (v) Find the heat produced in \( t = 0 \) to \( t = 5 \, s \)

1. **Using the formula for heat**:
\[ H = \int_0^5 \frac{E^2}{R} \, dt \]
Substitute \( E = 0.40 t + 0.40 \):
\[ H = \int_0^5 \frac{(0.40 t + 0.40)^2}{0.4} \, dt \]
Expanding the square:
\[ H = \int_0^5 \frac{(0.16 t^2 + 0.32 t + 0.16)}{0.4} \, dt = 0.4 \int_0^5 (0.16 t^2 + 0.32 t + 0.16) \, dt \]

2. **Calculating the integral**:
\[ H = 0.4 \left[ \frac{0.16 t^3}{3} + \frac{0.32 t^2}{2} + 0.16 t \right]_0^5 \]
Evaluating at the limits:
\[ H = 0.4 \left[ \frac{0.16 \cdot 125}{3} + \frac{0.32 \cdot 25}{2} + 0.16 \cdot 5 \right] \]
\[ = 0.4 \left[ \frac{20}{3} + 4 + 0.8 \right] = 0.4 \left[ \frac{20 + 12 + 2.4}{3} \right] = 0.4 \left[ \frac{34.4}{3} \right] \]
\[ = \frac{0.4 \times 34.4}{3} = \frac{13.76}{3} \approx 4.59 \, \text{J} \]

### Summary of Results:
- (i) Induced emf at \( t = 2 \, s \): \( -1.20 \, \text{V} \)
- (ii) Average induced emf from \( t = 0 \) to \( t = 5 \, s \): \( 1.40 \, \text{V} \)
- (iii) Charge passed through the loop: \( 17.5 \, \text{C} \)
- (iv) Average current: \( 3.5 \, \text{A} \)
- (v) Heat produced: \( \approx 4.59 \, \text{J} \)