A circular copper loop is placed perpendicular to a uniform magnetic field of 0.50 T. Due to external forces, the area of the loop decreases at a rate of `1.26 xx 10^(-3) m^(2)`/s. Determine the induced emf in the loop.

Options

A. `3.1xx10^(-4)V`
B. `1.2xx10^(-3)V`
C. `6.3xx10^(-4)V`
D. `7.9xx10^(-3)V`

Answer

To determine the induced emf in a circular copper loop placed perpendicular to a uniform magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Magnetic field strength, \( B = 0.50 \, \text{T} \)
- Rate of change of area, \( \frac{dA}{dt} = -1.26 \times 10^{-3} \, \text{m}^2/\text{s} \) (negative because the area is decreasing)

2. **Use the formula for induced emf:**
The induced emf (\( \mathcal{E} \)) in a loop can be calculated using Faraday's law of electromagnetic induction:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
where \( \Phi \) is the magnetic flux.

3. **Calculate the magnetic flux (\( \Phi \)):**
The magnetic flux through the loop is given by:
\[ \Phi = B \cdot A \]
Since the loop is perpendicular to the magnetic field, we can simplify this to:
\[ \Phi = B \cdot A \cdot \cos(0^\circ) = B \cdot A \]
where \( \cos(0^\circ) = 1 \).

4. **Differentiate the magnetic flux:**
Now, we differentiate the magnetic flux with respect to time:
\[ \frac{d\Phi}{dt} = B \cdot \frac{dA}{dt} \]

5. **Substitute the known values:**
Substitute \( B \) and \( \frac{dA}{dt} \) into the equation:
\[ \mathcal{E} = -\frac{d\Phi}{dt} = -B \cdot \frac{dA}{dt} \]
\[ \mathcal{E} = -0.50 \, \text{T} \cdot (-1.26 \times 10^{-3} \, \text{m}^2/\text{s}) \]

6. **Calculate the induced emf:**
\[ \mathcal{E} = 0.50 \cdot 1.26 \times 10^{-3} = 6.3 \times 10^{-4} \, \text{V} \]

### Final Answer:
The induced emf in the loop is \( \mathcal{E} = 6.3 \times 10^{-4} \, \text{V} \).

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