A square wire loop of side 30cm & wire cross section having diameter `4mm` is placed perpendicular to a magnetic field changing at the rate `0.2` T/s. Final induced current in the wire loop.
(Given: Resistivity of wire material is `(1.23xx10^(-8) ohm m)`

Options

A. `5.34xx10^2` A
B. `3.34xx10^2` A
C. `7.34xx10^2` A
D. `1.34xx10^2` A

Answer

To solve the problem of finding the final induced current in the square wire loop, we will follow these steps:

### Step 1: Calculate the Area of the Square Loop
The area \( A \) of a square loop can be calculated using the formula:
\[ A = \text{side}^2 \]
Given that the side length of the square loop is \( 30 \, \text{cm} = 0.3 \, \text{m} \):
\[ A = (0.3 \, \text{m})^2 = 0.09 \, \text{m}^2 \]

### Step 2: Calculate the Induced EMF
The induced electromotive force (EMF) can be calculated using Faraday's law of electromagnetic induction:
\[ \text{EMF} = -\frac{d\Phi}{dt} \]
Where \( \Phi \) is the magnetic flux given by \( \Phi = B \cdot A \). Since the magnetic field \( B \) is changing at a rate of \( \frac{dB}{dt} = 0.2 \, \text{T/s} \), we can express the induced EMF as:
\[ \text{EMF} = \frac{dB}{dt} \cdot A = 0.2 \, \text{T/s} \cdot 0.09 \, \text{m}^2 = 0.018 \, \text{V} = 18 \, \text{mV} \]

### Step 3: Calculate the Resistance of the Wire Loop
The resistance \( R \) of the wire loop can be calculated using the formula:
\[ R = \frac{\rho L}{A} \]
Where:
- \( \rho = 1.23 \times 10^{-8} \, \Omega \cdot \text{m} \) (resistivity of the wire material)
- \( L \) is the total length of the wire loop. Since it is a square loop, \( L = 4 \times \text{side} = 4 \times 0.3 \, \text{m} = 1.2 \, \text{m} \).
- The cross-sectional area \( a \) of the wire can be calculated using the diameter \( d = 4 \, \text{mm} = 0.004 \, \text{m} \):
\[ a = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.004}{2}\right)^2 = \pi \left(0.002\right)^2 = \pi \times 4 \times 10^{-6} \, \text{m}^2 \approx 1.2566 \times 10^{-5} \, \text{m}^2 \]

Now substituting these values into the resistance formula:
\[ R = \frac{(1.23 \times 10^{-8} \, \Omega \cdot \text{m}) \cdot (1.2 \, \text{m})}{1.2566 \times 10^{-5} \, \text{m}^2} \approx \frac{1.476 \times 10^{-8}}{1.2566 \times 10^{-5}} \approx 0.001175 \, \Omega \]

### Step 4: Calculate the Induced Current
Using Ohm's law, the induced current \( I \) can be calculated as:
\[ I = \frac{\text{EMF}}{R} \]
Substituting the values:
\[ I = \frac{0.018 \, \text{V}}{0.001175 \, \Omega} \approx 15.32 \, \text{A} \]

### Final Answer
The final induced current in the wire loop is approximately **15.32 A**.

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