A copper wire of length 40cm, diameter 2mm and resistivity `1.7xx10^(-8)Omega m` forms a square frame. If a uniform magnetic field B exists in a direction perpendicular to the plane of square frame and it changes at a steady rate `(dB)/(dt)=0.02` T/s, then find the current induced in the frame.

Options

A. `9.3xx10^(-2)` amp
B. `9.3xx10^(-1)` amp
C. `3.3xx10^(-2)` amp
D. `19.3xx10^(-2)` amp

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Convert Given Measurements
- Length of the copper wire: \( L = 40 \, \text{cm} = 0.4 \, \text{m} \)
- Diameter of the wire: \( d = 2 \, \text{mm} = 0.002 \, \text{m} \)
- Radius of the wire: \( r = \frac{d}{2} = \frac{0.002}{2} = 0.001 \, \text{m} \)
- Resistivity of copper: \( \rho = 1.7 \times 10^{-8} \, \Omega \, \text{m} \)

### Step 2: Calculate the Area of the Square Frame
Since the wire forms a square frame, the total length of the wire is equal to the perimeter of the square frame.
- Perimeter of the square frame: \( P = 4a \) where \( a \) is the side length.
- Therefore, \( a = \frac{L}{4} = \frac{0.4}{4} = 0.1 \, \text{m} \)

Now, calculate the area \( A \) of the square frame:
\[ A = a^2 = (0.1 \, \text{m})^2 = 0.01 \, \text{m}^2 \]

### Step 3: Calculate the Induced EMF
The induced EMF \( \mathcal{E} \) can be calculated using the formula:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
Where \( \Phi \) is the magnetic flux given by:
\[ \Phi = B \cdot A \]
Since \( A \) is constant, we have:
\[ \mathcal{E} = A \cdot \frac{dB}{dt} \]

Substituting the values:
- \( A = 0.01 \, \text{m}^2 \)
- \( \frac{dB}{dt} = 0.02 \, \text{T/s} \)

Thus,
\[ \mathcal{E} = 0.01 \cdot 0.02 = 2 \times 10^{-4} \, \text{V} \]

### Step 4: Calculate the Resistance of the Wire
The resistance \( R \) of the wire can be calculated using the formula:
\[ R = \frac{\rho L}{A_w} \]
Where \( A_w \) is the cross-sectional area of the wire given by:
\[ A_w = \pi r^2 \]

Calculating \( A_w \):
\[ A_w = \pi (0.001)^2 = \pi \times 10^{-6} \approx 3.14 \times 10^{-6} \, \text{m}^2 \]

Now substituting the values into the resistance formula:
\[ R = \frac{(1.7 \times 10^{-8}) \cdot (0.4)}{3.14 \times 10^{-6}} \]
\[ R \approx \frac{6.8 \times 10^{-9}}{3.14 \times 10^{-6}} \approx 2.16 \times 10^{-3} \, \Omega \]

### Step 5: Calculate the Induced Current
Using Ohm's law, the induced current \( I \) can be calculated as:
\[ I = \frac{\mathcal{E}}{R} \]
Substituting the values:
\[ I = \frac{2 \times 10^{-4}}{2.16 \times 10^{-3}} \]
\[ I \approx 9.26 \times 10^{-2} \, \text{A} \]

### Final Answer
The induced current in the frame is approximately:
\[ I \approx 0.0926 \, \text{A} \]

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