A coil of area 500 `cm^(2)` having 1000 turns is put perpendicular to a magnetic field of intensity `4xx10^(-5)`T. if it is rotated by `180^(@)` in 0.1 s, the induced emf produced is

Answer

To solve the problem step by step, we will use the formula for induced electromotive force (emf) in a coil due to a change in magnetic flux.

### Given Data:
- Area of the coil, \( A = 500 \, \text{cm}^2 = 500 \times 10^{-4} \, \text{m}^2 = 0.05 \, \text{m}^2 \)
- Number of turns, \( N = 1000 \)
- Magnetic field intensity, \( B = 4 \times 10^{-5} \, \text{T} \)
- Angle of rotation, \( \Delta \theta = 180^\circ \)
- Time for rotation, \( \Delta t = 0.1 \, \text{s} \)

### Step 1: Calculate the initial magnetic flux (\( \Phi_i \))
The magnetic flux (\( \Phi \)) through the coil is given by:
\[ \Phi = N \cdot B \cdot A \cdot \cos(\theta) \]
Initially, the coil is perpendicular to the magnetic field, so \( \theta = 0^\circ \) and \( \cos(0) = 1 \).

\[ \Phi_i = N \cdot B \cdot A \cdot \cos(0) = 1000 \cdot (4 \times 10^{-5}) \cdot (0.05) \cdot 1 \]
\[ \Phi_i = 1000 \cdot 4 \times 10^{-5} \cdot 0.05 = 0.002 \, \text{Wb} \]

### Step 2: Calculate the final magnetic flux (\( \Phi_f \))
After rotating the coil by \( 180^\circ \), the angle becomes \( 180^\circ \) and \( \cos(180) = -1 \).

\[ \Phi_f = N \cdot B \cdot A \cdot \cos(180) = 1000 \cdot (4 \times 10^{-5}) \cdot (0.05) \cdot (-1) \]
\[ \Phi_f = 1000 \cdot 4 \times 10^{-5} \cdot 0.05 \cdot (-1) = -0.002 \, \text{Wb} \]

### Step 3: Calculate the change in magnetic flux (\( \Delta \Phi \))
The change in magnetic flux is given by:
\[ \Delta \Phi = \Phi_f - \Phi_i \]
\[ \Delta \Phi = -0.002 - 0.002 = -0.004 \, \text{Wb} \]

### Step 4: Calculate the induced emf (\( E \))
The average induced emf can be calculated using the formula:
\[ E = -\frac{\Delta \Phi}{\Delta t} \]
Substituting the values:
\[ E = -\frac{-0.004}{0.1} = 0.04 \, \text{V} = 40 \, \text{mV} \]

### Final Answer:
The induced emf produced is \( 40 \, \text{mV} \).

---