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 find 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:
\[ A = L^2 \]
where \( L \) is the side length of the square. Given that the side length is \( 30 \, \text{cm} \) (which is \( 0.3 \, \text{m} \)):
\[ A = (0.3 \, \text{m})^2 = 0.09 \, \text{m}^2 \]

### Step 2: Calculate the Rate of Change of Magnetic Flux
The rate of change of magnetic flux \( \frac{d\Phi}{dt} \) is given by:
\[ \frac{d\Phi}{dt} = \frac{dB}{dt} \cdot A \]
Given \( \frac{dB}{dt} = 0.2 \, \text{T/s} \):
\[ \frac{d\Phi}{dt} = 0.2 \, \text{T/s} \cdot 0.09 \, \text{m}^2 = 0.018 \, \text{Wb/s} \]

### Step 3: Calculate the Induced EMF
Using Faraday's law of electromagnetic induction, the induced EMF \( \mathcal{E} \) is equal to the rate of change of magnetic flux:
\[ \mathcal{E} = \frac{d\Phi}{dt} = 0.018 \, \text{V} \]

### Step 4: Calculate the Resistance of the Wire Loop
The resistance \( R \) of the wire can be calculated using the formula:
\[ R = \frac{\rho L}{A_c} \]
where \( \rho \) is the resistivity, \( L \) is the total length of the wire, and \( A_c \) is the cross-sectional area of the wire.

1. **Calculate the total length \( L \)**:
Since it is a square loop, the total length \( L \) is:
\[ L = 4 \times 0.3 \, \text{m} = 1.2 \, \text{m} \]

2. **Calculate the cross-sectional area \( A_c \)**:
The diameter of the wire is \( 4 \, \text{mm} = 0.004 \, \text{m} \), so the radius \( r \) is:
\[ r = \frac{0.004}{2} = 0.002 \, \text{m} \]
The cross-sectional area \( A_c \) is:
\[ A_c = \pi r^2 = \pi (0.002)^2 = \pi \times 4 \times 10^{-6} \approx 1.25664 \times 10^{-5} \, \text{m}^2 \]

3. **Calculate the resistance \( R \)**:
Given \( \rho = 1.23 \times 10^{-8} \, \Omega \cdot \text{m} \):
\[ R = \frac{(1.23 \times 10^{-8}) \times (1.2)}{1.25664 \times 10^{-5}} \approx \frac{1.476 \times 10^{-8}}{1.25664 \times 10^{-5}} \approx 0.001176 \, \Omega \]

### 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:
\[ I = \frac{0.018}{0.001176} \approx 15.32 \, \text{A} \]

### Final Answer
The final induced current in the wire loop is approximately \( 15.32 \, \text{A} \).