a coil of `1200` turns and mean area of `500 cm^(2)` is held perpendicular to a uniform magnetic field of induction `4 xx 10^(-4) T`. The resistance of the coil is `20` ohms. When the coil is rotated through `180^(@)` in the magnetic field in `0.1` seconds the average electric current (in `mA`) induced is :

Options

A. `12`
B. `24`
C. `36`
D. `48`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the initial and final magnetic flux
The magnetic flux (Φ) through the coil is given by the formula:
\[ \Phi = n \cdot B \cdot A \cdot \cos(\theta) \]
Where:
- \( n = 1200 \) (number of turns)
- \( B = 4 \times 10^{-4} \, T \) (magnetic field induction)
- \( A = 500 \, cm^2 = 500 \times 10^{-4} \, m^2 = 5 \times 10^{-2} \, m^2 \) (area in square meters)
- \( \theta \) is the angle between the magnetic field and the normal to the surface of the coil.

Initially, the coil is perpendicular to the magnetic field, so \( \theta = 0^\circ \):
\[ \Phi_i = n \cdot B \cdot A \cdot \cos(0) = 1200 \cdot (4 \times 10^{-4}) \cdot (5 \times 10^{-2}) \cdot 1 \]
Calculating this:
\[ \Phi_i = 1200 \cdot 4 \cdot 5 \times 10^{-6} = 24 \times 10^{-3} \, Wb \]

After rotating the coil through \( 180^\circ \), \( \theta = 180^\circ \):
\[ \Phi_f = n \cdot B \cdot A \cdot \cos(180) = 1200 \cdot (4 \times 10^{-4}) \cdot (5 \times 10^{-2}) \cdot (-1) \]
Calculating this:
\[ \Phi_f = -24 \times 10^{-3} \, Wb \]

### Step 2: Calculate the change in magnetic flux
The change in magnetic flux (ΔΦ) is given by:
\[ \Delta \Phi = \Phi_f - \Phi_i = -24 \times 10^{-3} - 24 \times 10^{-3} = -48 \times 10^{-3} \, Wb \]

### Step 3: Calculate the induced EMF (ε)
According to Faraday's law of electromagnetic induction, the induced EMF (ε) is given by:
\[ \epsilon = -\frac{\Delta \Phi}{\Delta t} \]
Where \( \Delta t = 0.1 \, s \):
\[ \epsilon = -\frac{-48 \times 10^{-3}}{0.1} = \frac{48 \times 10^{-3}}{0.1} = 0.48 \, V \]

### Step 4: Calculate the induced current (I)
Using Ohm's law, the induced current (I) can be calculated as:
\[ I = \frac{\epsilon}{R} \]
Where \( R = 20 \, \Omega \):
\[ I = \frac{0.48}{20} = 0.024 \, A \]

### Step 5: Convert the current to milliampere
To convert amperes to milliamperes:
\[ I = 0.024 \, A = 24 \, mA \]

### Final Answer
The average electric current induced is **24 mA**.

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