A square loop of wire with side length 10 cm is placed at angle of `45^(@)` with a magnetic field that changes uniformly from 0.1 T to zero in 0.7s. The induced current in the loop (its resistance is `1Omega`) is

Options

A. 1.0 mA
B. 2.5 mA
C. 3.5 mA
D. 4.0 mA

Explanation

To solve the problem step by step, we will follow the principles of electromagnetic induction and use the relevant formulas.

### Step 1: Identify the given parameters
- Side length of the square loop, \( L = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Angle with the magnetic field, \( \theta = 45^\circ \)
- Initial magnetic field, \( B_i = 0.1 \, \text{T} \)
- Final magnetic field, \( B_f = 0 \, \text{T} \)
- Time taken for the change, \( t = 0.7 \, \text{s} \)
- Resistance of the loop, \( R = 1 \, \Omega \)

### Step 2: Calculate the area of the square loop
The area \( A \) of the square loop is given by:
\[ A = L^2 = (0.1 \, \text{m})^2 = 0.01 \, \text{m}^2 \]

### Step 3: Calculate the change in magnetic flux
The magnetic flux \( \Phi \) through the loop is given by:
\[ \Phi = B \cdot A \cdot \cos(\theta) \]
- Initial flux \( \Phi_i \):
\[ \Phi_i = B_i \cdot A \cdot \cos(45^\circ) = 0.1 \cdot 0.01 \cdot \frac{1}{\sqrt{2}} = \frac{0.001}{\sqrt{2}} \, \text{Wb} \]
- Final flux \( \Phi_f \):
\[ \Phi_f = B_f \cdot A \cdot \cos(45^\circ) = 0 \cdot 0.01 \cdot \frac{1}{\sqrt{2}} = 0 \, \text{Wb} \]

### Step 4: Calculate the change in flux
The change in flux \( \Delta \Phi \) is:
\[ \Delta \Phi = \Phi_f - \Phi_i = 0 - \frac{0.001}{\sqrt{2}} = -\frac{0.001}{\sqrt{2}} \, \text{Wb} \]

### Step 5: Calculate the induced EMF
According to Faraday's law of electromagnetic induction, the induced EMF \( \mathcal{E} \) is given by:
\[ \mathcal{E} = -\frac{\Delta \Phi}{t} = -\frac{-\frac{0.001}{\sqrt{2}}}{0.7} \]
\[ \mathcal{E} = \frac{0.001}{0.7 \sqrt{2}} \, \text{V} \]

### Step 6: Calculate the induced current
Using Ohm's law, the induced current \( I \) can be calculated as:
\[ I = \frac{\mathcal{E}}{R} = \frac{\frac{0.001}{0.7 \sqrt{2}}}{1} \]
\[ I = \frac{0.001}{0.7 \sqrt{2}} \, \text{A} \]

### Step 7: Numerical calculation
Calculating the numerical value:
\[ \sqrt{2} \approx 1.414 \]
\[ I \approx \frac{0.001}{0.7 \times 1.414} \approx \frac{0.001}{0.9898} \approx 0.00101 \, \text{A} \approx 1.01 \, \text{mA} \]

### Final Answer
The induced current in the loop is approximately \( 1.01 \, \text{mA} \).

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