A coil of 1500 turns and mean area of `500 cm^2` is held perpendicular to a uniform magnetic field of induction `2 xx 10^(-4)` T . The resistance of the coil is 40 ohms. When the coil is rotated through `180^@` in the magnetic field in 0.2 seconds the average electric current (in mA) induced is

Options

A. 1.2
B. 2.4
C. 3.7
D. 4.3

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Convert the area from cm² to m²
The area of the coil is given as \(500 \, \text{cm}^2\). We need to convert this to square meters.

\[ \text{Area} = 500 \, \text{cm}^2 = 500 \times 10^{-4} \, \text{m}^2 = 0.05 \, \text{m}^2 \]

### Step 2: Calculate the initial magnetic flux
The magnetic flux (\(\Phi\)) through the coil when it is perpendicular to the magnetic field is given by:

\[ \Phi = n \cdot B \cdot A \cdot \cos(\theta) \]

Where:
- \(n = 1500\) (number of turns)
- \(B = 2 \times 10^{-4} \, \text{T}\) (magnetic field induction)
- \(A = 0.05 \, \text{m}^2\) (area)
- \(\theta = 0\) (initial angle)

Since \(\cos(0) = 1\):

\[ \Phi_{\text{initial}} = 1500 \cdot (2 \times 10^{-4}) \cdot 0.05 \cdot 1 \]

Calculating this gives:

\[ \Phi_{\text{initial}} = 1500 \cdot 2 \cdot 0.05 \times 10^{-4} = 1500 \cdot 0.1 \times 10^{-4} = 150 \times 10^{-4} = 1.5 \times 10^{-2} \, \text{Wb} \]

### Step 3: Calculate the final magnetic flux after rotation
After rotating the coil through \(180^\circ\), the angle becomes \(180^\circ\) and \(\cos(180^\circ) = -1\):

\[ \Phi_{\text{final}} = n \cdot B \cdot A \cdot \cos(180^\circ) = 1500 \cdot (2 \times 10^{-4}) \cdot 0.05 \cdot (-1) \]

Calculating this gives:

\[ \Phi_{\text{final}} = -1.5 \times 10^{-2} \, \text{Wb} \]

### Step 4: Calculate the change in magnetic flux
The change in magnetic flux (\(\Delta \Phi\)) is given by:

\[ \Delta \Phi = \Phi_{\text{final}} - \Phi_{\text{initial}} = (-1.5 \times 10^{-2}) - (1.5 \times 10^{-2}) = -3.0 \times 10^{-2} \, \text{Wb} \]

### Step 5: Calculate the induced EMF
The induced EMF (\(E\)) can be calculated using Faraday's law of electromagnetic induction:

\[ E = -\frac{\Delta \Phi}{\Delta t} \]

Where \(\Delta t = 0.2 \, \text{s}\):

\[ E = -\frac{-3.0 \times 10^{-2}}{0.2} = \frac{3.0 \times 10^{-2}}{0.2} = 0.15 \, \text{V} = 150 \, \text{mV} \]

### Step 6: Calculate the induced current
Using Ohm's law, the induced current (\(I\)) can be calculated as:

\[ I = \frac{E}{R} \]

Where \(R = 40 \, \Omega\):

\[ I = \frac{0.15}{40} = 0.00375 \, \text{A} = 3.75 \, \text{mA} \]

### Final Answer
The average electric current induced is \(3.75 \, \text{mA}\).

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