A conducting circular loop made of a thin wire has area $$3.5 \times 10^{-2}$$ m$$^2$$ and resistance 10 $$\Omega$$. It is placed perpendicular to a time-dependent magnetic field $$B(t) = 0.4 \; T \sin 50\pi t$$. The field is uniform in space. Then the net charge flowing through the loop during $$t = 0$$ s and $$t = 10$$ ms is close to:

We have a circular conducting loop whose plane is kept perpendicular to the magnetic field, so the magnetic flux through the loop is simply the product of the field magnitude and the loop area.

The area is given as $$A = 3.5 \times 10^{-2}\,{\rm m^2}$$ and the magnetic field varies with time as $$B(t) = 0.4\,{\rm T}\,\sin(50\pi t)$$, where the time $$t$$ is in seconds. Hence the magnetic flux $$\Phi(t)$$ through the loop is

$$\Phi(t) \;=\; B(t)\,A \;=\; \bigl(0.4\sin 50\pi t\bigr)\times \bigl(3.5\times10^{-2}\bigr) \;=\; 0.014\,\sin 50\pi t\;{\rm Wb}.$$

According to Faraday’s law of electromagnetic induction, the induced emf $$\mathcal{E}(t)$$ in the loop is related to the time-rate of change of flux by

$$\mathcal{E}(t) = -\dfrac{d\Phi(t)}{dt}.$$

The loop has resistance $$R = 10\,\Omega$$. Using Ohm’s law, the induced current is

$$I(t) = \dfrac{\mathcal{E}(t)}{R} = -\dfrac{1}{R}\,\dfrac{d\Phi(t)}{dt}.$$

The total charge $$Q$$ that passes through any cross-section of the wire from time $$t=0$$ to $$t = 10\,{\rm ms} = 0.01\,{\rm s}$$ is obtained by integrating the current:

$$Q = \int_{0}^{0.01} I(t)\,dt = -\dfrac{1}{R}\int_{0}^{0.01}\dfrac{d\Phi(t)}{dt}\,dt = -\dfrac{1}{R}\,\bigl[\Phi(t)\bigr]_{0}^{0.01}.$$

Carrying out the substitution of the flux values, we first evaluate $$\Phi(t)$$ at both limits:

At $$t = 0$$:

$$\Phi(0) = 0.014\,\sin\bigl(50\pi \times 0\bigr) = 0.014\,\sin 0 = 0.$$

At $$t = 0.01\,{\rm s}$$:

$$50\pi t = 50\pi \times 0.01 = 0.5\pi = \dfrac{\pi}{2},$$

so $$\sin\!\Bigl(\dfrac{\pi}{2}\Bigr) = 1,$$ and

$$\Phi(0.01) = 0.014 \times 1 = 0.014\,{\rm Wb}.$$

The change in flux is therefore

$$\Delta\Phi = \Phi(0.01) - \Phi(0) = 0.014 - 0 = 0.014\,{\rm Wb}.$$

Substituting this into the expression for charge, we get

$$Q = -\dfrac{1}{10}\,\bigl[\,0.014 - 0\,\bigr] = -\dfrac{0.014}{10} = -0.0014\,{\rm C}.$$

The negative sign only indicates the direction of flow; the magnitude of charge that actually passes through the loop is

$$|Q| = 0.0014\,{\rm C} = 1.4 \times 10^{-3}\,{\rm C} = 1.4\,{\rm mC}.$$

Hence, the correct answer is Option A.