Transverse Voltage and Resistance of Semicircular Metallic Strip

To determine the correct statements, we analyze each property of the semicircular metallic strip step-by-step.

1. Calculation of Current $I$ (Option A)

Consider a thin semicircular element of radius $r$ and thickness $dr$ along the radial direction.

• The length of this elemental strip is $L = \pi r$.
• The cross-sectional area through which the current flows is $dA = t \, dr$.

The resistance $dR$ of this elemental strip is: $dR = \rho \frac{\pi r}{t \, dr}$

Since all such elemental strips are connected in parallel across the potential difference $V_0$, the net conductance $G = \frac{1}{R}$ of the entire strip is obtained by integrating the differential conductances $dG = \frac{1}{dR}$: $G = \int_{R_1}^{R_2} \frac{t \, dr}{\pi \rho r} = \frac{t}{\pi \rho} \ln\left(\frac{R_2}{R_1}\right)$

Therefore, the total current $I$ passing through the strip is: $I = V_0 G = \frac{V_0 t}{\pi \rho} \ln\left(\frac{R_2}{R_1}\right)$

This confirms that Option A is correct.

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2. Potential of Inner vs. Outer Surfaces (Options B and C)

The electrons drift along concentric circular paths to form the electric current. For an electron of mass $m$ and charge $-e$ moving with drift velocity $v(r)$ along a circle of radius $r$, the required centripetal acceleration is directed radially inward towards the center: $a_c = \frac{v(r)^2}{r}$

This centripetal acceleration must be provided by a radial electric force $-e E_r$: $-e E_r = -\frac{m v(r)^2}{r} \implies E_r = \frac{m v(r)^2}{e r} > 0$

Since $E_r > 0$, the radial electric field points radially outward (from $R_1$ to $R_2$).

The electric field is related to the electric potential $V(r)$ by: $E_r = -\frac{dV}{dr} \implies \frac{dV}{dr} = -\frac{m v(r)^2}{e r} < 0$

Because $\frac{dV}{dr} < 0$, the potential decreases as $r$ increases. Consequently, the potential at the outer surface $V(R_2)$ is lower than the potential at the inner surface $V(R_1)$: $V(R_2) < V(R_1)$

Thus, the outer surface is at a lower voltage than the inner surface. This confirms that Option C is correct and Option B is incorrect.

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3. Dependence of Transverse Voltage $\Delta V$ on Current $I$ (Option D)

The tangential electric field at radius $r$ is: $E_\theta(r) = \frac{V_0}{\pi r}$

According to Ohm's law, the current density $J(r)$ is: $J(r) = \frac{E_\theta(r)}{\rho} = \frac{V_0}{\pi \rho r}$

Using $J(r) = n e v(r)$, the drift velocity of electrons is: $v(r) = \frac{V_0}{n e \pi \rho r}$

Since $V_0 \propto I$ (from Option A), we have $v(r) \propto I$.

The transverse potential difference $\Delta V$ between the inner and outer boundaries is: $\Delta V = V(R_1) - V(R_2) = \int_{R_1}^{R_2} E_r \, dr = \int_{R_1}^{R_2} \frac{m v(r)^2}{e r} \, dr$

Since $v(r) \propto I$, it follows that $v(r)^2 \propto I^2$. Substituting this into the integral gives: $\Delta V \propto I^2$

This confirms that Option D is correct.

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Conclusion

The correct options are A, C, and D.