Induced EMF in Circular Loop from Straight Wire Current

To determine the induced electromotive force ($\text{emf}$) in the circular wire loop, we apply Faraday's Law of Electromagnetic Induction.

1. Magnetic Field due to the Straight Wire

Consider an infinitely long straight wire carrying a current $I(t)$ placed along the diameter of a circular loop of radius $R$. Set up a coordinate system in the plane of the circular loop with the straight wire lying along the $x$-axis ($y = 0$).

The magnetic field $\vec{B}$ at a distance $y$ from the straight wire is given by Ampere's Law: $\vec{B}(y) = \frac{\mu_0 I(t)}{2\pi y} \hat{k}$ where:

• For $y > 0$ (upper half-circle), $\vec{B}$ points out of the plane ($+\hat{k}$).
• For $y < 0$ (lower half-circle), $\vec{B}$ points into the plane ($-\hat{k}$).

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2. Calculation of Net Magnetic Flux

The total magnetic flux $\Phi_{\text{net}}$ passing through the circular loop is the surface integral of the magnetic field over the area $S$ bounded by the loop: $\Phi_{\text{net}} = \iint_S \vec{B} \cdot d\vec{A}$

Dividing the circular area into two symmetric semi-circular halves $S_1$ (for $y > 0$) and $S_2$ (for $y < 0$):

$\Phi_{\text{net}} = \iint_{S_1} \vec{B} \cdot d\vec{A} + \iint_{S_2} \vec{B} \cdot d\vec{A}$

Taking the area vector $d\vec{A} = dA \, \hat{k}$ outward normal to the plane:

1. Flux through upper semicircle ($S_1$): $\Phi_1 = \iint_{S_1} \left(\frac{\mu_0 I(t)}{2\pi y}\right) dA = +\Phi_0(t)$

2. Flux through lower semicircle ($S_2$): $\Phi_2 = \iint_{S_2} \left(-\frac{\mu_0 I(t)}{2\pi |y|}\right) dA = -\Phi_0(t)$

Due to the complete spatial symmetry about the diameter ($x$-axis): $\Phi_{\text{net}} = \Phi_1 + \Phi_2 = \Phi_0(t) - \Phi_0(t) = 0$

Thus, the total magnetic flux through the loop is zero at all times $t$, regardless of the current $I(t)$.

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3. Calculation of Induced EMF

By Faraday's Law of Electromagnetic Induction, the induced $\text{emf}$ ($\mathcal{E}$) in the loop is given by: $\mathcal{E} = -\frac{d\Phi_{\text{net}}}{dt}$

Since $\Phi_{\text{net}}(t) = 0$ identically for all $t$: $\mathcal{E} = -\frac{d(0)}{dt} = 0$

• Case 1: If the current $I$ is constant ($I = \text{constant}$), then $\Phi_{\text{net}} = 0 \implies \mathcal{E} = 0$.
• Case 2: If the current decreases at a uniform rate ($\frac{dI}{dt} = \text{constant} < 0$), the total flux $\Phi_{\text{net}}$ remains identically zero at all instants, so $\mathcal{E} = 0$.

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Conclusion

• Statement (A) is correct.
• Statement (C) is correct.

Correct Options: A, C