To solve the problem of finding the net charge flowing through a conducting circular loop placed in a time-varying magnetic field, we can follow these steps:
### Step-by-Step Solution:
1. **Identify the Given Values:**
- Area of the loop, \( A = 2.5 \times 10^{-3} \, m^2 \)
- Resistance of the loop, \( R = 10 \, \Omega \)
- Magnetic field, \( B(t) = 0.6 \sin(50 \pi t) \, T \)
2. **Calculate the Magnetic Flux:**
The magnetic flux \( \Phi \) through the loop is given by:
\[
\Phi = B \cdot A
\]
Substituting the expression for \( B(t) \):
\[
\Phi(t) = A \cdot B(t) = A \cdot (0.6 \sin(50 \pi t))
\]
Therefore,
\[
\Phi(t) = 2.5 \times 10^{-3} \cdot 0.6 \sin(50 \pi t) = 1.5 \times 10^{-3} \sin(50 \pi t) \, Wb
\]
3. **Determine the Induced EMF:**
The induced electromotive force (EMF) \( E \) in the loop can be calculated using Faraday's law of electromagnetic induction:
\[
E = -\frac{d\Phi}{dt}
\]
To find \( \frac{d\Phi}{dt} \):
\[
\frac{d\Phi}{dt} = 1.5 \times 10^{-3} \cdot \frac{d}{dt}(\sin(50 \pi t)) = 1.5 \times 10^{-3} \cdot (50 \pi \cos(50 \pi t))
\]
Thus,
\[
E = -1.5 \times 10^{-3} \cdot (50 \pi \cos(50 \pi t)) = -75 \pi \times 10^{-3} \cos(50 \pi t) \, V
\]
4. **Calculate the Current:**
Using Ohm's law, the current \( I \) flowing through the loop is given by:
\[
I = \frac{E}{R}
\]
Substituting the expression for \( E \):
\[
I = \frac{-75 \pi \times 10^{-3} \cos(50 \pi t)}{10} = -7.5 \pi \times 10^{-3} \cos(50 \pi t) \, A
\]
5. **Calculate the Charge Flowing through the Loop:**
The charge \( Q \) that flows through the loop during the time interval from \( t = 0 \) to \( t = 10 \, ms \) can be calculated by integrating the current over this time period:
\[
Q = \int_{0}^{10 \times 10^{-3}} I \, dt = \int_{0}^{10 \times 10^{-3}} -7.5 \pi \times 10^{-3} \cos(50 \pi t) \, dt
\]
The integral of \( \cos(50 \pi t) \) is:
\[
\int \cos(50 \pi t) \, dt = \frac{1}{50 \pi} \sin(50 \pi t)
\]
Thus,
\[
Q = -7.5 \pi \times 10^{-3} \left[ \frac{1}{50 \pi} \sin(50 \pi t) \right]_{0}^{10 \times 10^{-3}}
\]
Evaluating the limits:
\[
Q = -7.5 \times 10^{-3} \left[ \frac{1}{50} \sin(0.5) - 0 \right] = -7.5 \times 10^{-3} \cdot \frac{1}{50} \sin(0.5)
\]
Using \( \sin(0.5) \approx 0.4794 \):
\[
Q \approx -7.5 \times 10^{-3} \cdot \frac{0.4794}{50} \approx -7.5 \times 10^{-3} \cdot 0.009588 \approx -7.2 \times 10^{-5} \, C
\]
Converting to millicoulombs:
\[
Q \approx -0.072 \, mC
\]
### Final Answer:
The net charge flowing through the loop during the time interval from \( t = 0 \) to \( t = 10 \, ms \) is approximately \( 0.072 \, mC \).
