To solve the problem step by step, we will follow these steps:
### Step 1: Calculate the Area of the Square Loop
The area \( A \) of a square loop is given by the formula:
\[
A = \text{side}^2
\]
Given the side of the square loop is 12 cm, we first convert it to meters:
\[
\text{side} = 12 \text{ cm} = 0.12 \text{ m}
\]
Now, calculate the area:
\[
A = (0.12 \, \text{m})^2 = 0.0144 \, \text{m}^2
\]
**Hint:** Remember to convert centimeters to meters when calculating area in SI units.
### Step 2: Determine the Initial Magnetic Flux
The initial magnetic flux \( \Phi_1 \) can be calculated using the formula:
\[
\Phi_1 = B_1 \cdot A \cdot \cos(\theta)
\]
Where:
- \( B_1 = 0.10 \, \text{T} \) (initial magnetic field)
- \( A = 0.0144 \, \text{m}^2 \) (area calculated in Step 1)
- \( \theta = 45^\circ \) (angle between the magnetic field and the normal to the loop)
Now, calculate \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \approx 0.707 \):
\[
\Phi_1 = 0.10 \cdot 0.0144 \cdot 0.707 \approx 0.00102 \, \text{Wb}
\]
**Hint:** Use the cosine of the angle to find the component of the magnetic field that is perpendicular to the area.
### Step 3: Determine the Final Magnetic Flux
The final magnetic flux \( \Phi_2 \) when the magnetic field is zero is:
\[
\Phi_2 = 0 \, \text{Wb}
\]
**Hint:** When the magnetic field is zero, the flux through the loop is also zero.
### Step 4: Calculate the Change in Magnetic Flux
The change in magnetic flux \( \Delta \Phi \) is given by:
\[
\Delta \Phi = \Phi_2 - \Phi_1 = 0 - 0.00102 = -0.00102 \, \text{Wb}
\]
**Hint:** The change in flux is simply the final flux minus the initial flux.
### Step 5: Calculate the Induced EMF
The induced electromotive force (EMF) \( E \) can be calculated using Faraday's law:
\[
E = -\frac{\Delta \Phi}{\Delta t}
\]
Where \( \Delta t = 0.6 \, \text{s} \):
\[
E = -\frac{-0.00102}{0.6} \approx 0.0017 \, \text{V}
\]
**Hint:** The negative sign indicates the direction of the induced EMF, but we are interested in the magnitude.
### Step 6: Calculate the Current
Using Ohm's Law, the current \( I \) can be calculated as:
\[
I = \frac{E}{R}
\]
Where \( R = 0.6 \, \Omega \):
\[
I = \frac{0.0017}{0.6} \approx 0.00283 \, \text{A} \text{ or } 2.83 \, \text{mA}
\]
**Hint:** Make sure to use the resistance in ohms to find the current in amperes.
### Final Answer
The magnitude of the current during this time interval is approximately \( 2.83 \, \text{mA} \).
