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Step-by-Step Solution
Step 1: Identify the Given Data
• Area of the loop, $A = 3.5 \times 10^{-3}\,\mathrm{m}^2$
• Resistance of the loop, $R = 10\,\Omega$
• Magnetic field as a function of time, $B(t) = (0.4\,\mathrm{T})\,\sin(50\pi t)$
• We need the net charge flowing through the loop from $t=0\,\mathrm{s}$ to $t=10\,\mathrm{ms}$ (i.e., $t = 10 \times 10^{-3}\,\mathrm{s}$).
Step 2: Determine the Magnetic Field at the Specified Times
1. At $t = 0$:
\[
B(0) = 0.4 \,\mathrm{T} \cdot \sin(50\pi \cdot 0) = 0.4 \,\mathrm{T} \cdot 0 = 0 \,\mathrm{T}.
\]
2. At $t = 10 \times 10^{-3}\,\mathrm{s}$:
\[
B(10\,\mathrm{ms})
= 0.4 \,\mathrm{T} \cdot \sin\bigl(50\pi \times 10\times 10^{-3}\bigr)
= 0.4 \,\mathrm{T} \cdot \sin\bigl(50\pi \times 10^{-2}\bigr)
= 0.4 \,\mathrm{T} \cdot \sin\left(\frac{\pi}{2}\right) = 0.4 \,\mathrm{T}.
\]
Step 3: Calculate the Change in Magnetic Flux
Magnetic flux through the loop at time $t$ is given by $\phi(t) = A \, B(t)$ (since the field is perpendicular to the loop). The change in flux, $\Delta \phi$, between $t=0$ and $t=10\,\mathrm{ms}$ is:
\[
\Delta \phi = A \bigl[B(10\,\mathrm{ms}) - B(0)\bigr]
= 3.5 \times 10^{-3} \,\mathrm{m}^2 \,\bigl(0.4 - 0 \bigr)
= 3.5 \times 10^{-3} \times 0.4
= 1.4 \times 10^{-3}\,\mathrm{Wb}.
\]
Step 4: Relate Change in Flux to Net Charge
By Faraday’s law of induction and the concept of total charge flow, the net charge $q$ passing through the conductor is:
\[
q = \frac{\Delta \phi}{R}.
\]
Substituting $\Delta \phi = 1.4 \times 10^{-3}\,\mathrm{Wb}$ and $R = 10\,\Omega$:
\[
q = \frac{1.4 \times 10^{-3}}{10}
= 0.14 \times 10^{-3}\,\mathrm{C}
= 0.14 \,\mathrm{mC}.
\]
Step 5: Final Answer
The net charge flowing through the loop is approximately 0.14 mC.
