To solve the problem, we will follow these steps:
### Step 1: Understand the parameters given in the problem
- Side of the square loop, \( a = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Resistance of the loop, \( R = 0.5 \, \Omega \)
- Initial magnetic field, \( B_i = 0.1 \, \text{T} \)
- Final magnetic field, \( B_f = 0 \, \text{T} \)
- Time interval for the change in magnetic field, \( \Delta t = 0.7 \, \text{s} \)
### Step 2: Calculate the area of the square loop
The area \( A \) of the square loop is given by:
\[
A = a^2 = (0.1 \, \text{m})^2 = 0.01 \, \text{m}^2
\]
### Step 3: Calculate the initial magnetic flux
The magnetic flux \( \Phi \) through the loop is given by:
\[
\Phi = B \cdot A \cdot \cos(\theta)
\]
Since the magnetic field is in the North-East direction and the loop is in the East-West plane, the angle \( \theta \) between the magnetic field and the area vector is \( 45^\circ \). Thus, \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \).
Calculating the initial flux:
\[
\Phi_i = B_i \cdot A \cdot \cos(45^\circ) = 0.1 \, \text{T} \cdot 0.01 \, \text{m}^2 \cdot \frac{1}{\sqrt{2}} = 0.000707 \, \text{Wb} \, (\text{Weber})
\]
### Step 4: Calculate the final magnetic flux
Since the final magnetic field \( B_f = 0 \, \text{T} \), the final flux is:
\[
\Phi_f = B_f \cdot A \cdot \cos(45^\circ) = 0 \cdot 0.01 \cdot \frac{1}{\sqrt{2}} = 0 \, \text{Wb}
\]
### Step 5: Calculate the change in magnetic flux
The change in magnetic flux \( \Delta \Phi \) is given by:
\[
\Delta \Phi = \Phi_f - \Phi_i = 0 - 0.000707 \, \text{Wb} = -0.000707 \, \text{Wb}
\]
### Step 6: Calculate the induced EMF
According to Faraday's law of electromagnetic induction, the induced EMF \( \mathcal{E} \) is given by:
\[
\mathcal{E} = -\frac{\Delta \Phi}{\Delta t}
\]
Substituting the values:
\[
\mathcal{E} = -\frac{-0.000707 \, \text{Wb}}{0.7 \, \text{s}} = \frac{0.000707}{0.7} \approx 0.00101 \, \text{V} \, (\text{or } 1.01 \, \text{mV})
\]
### Step 7: Calculate the induced current
Using Ohm's law, the induced current \( I \) can be calculated as:
\[
I = \frac{\mathcal{E}}{R}
\]
Substituting the values:
\[
I = \frac{0.00101 \, \text{V}}{0.5 \, \Omega} \approx 0.00202 \, \text{A} \, (\text{or } 2.02 \, \text{mA})
\]
### Final Results
- Induced EMF: \( \mathcal{E} \approx 1.01 \, \text{mV} \)
- Induced Current: \( I \approx 2.02 \, \text{mA} \)
