Shown in the figure is a semicircular metallic strip that has thickness t and resistivity rho. Its i

Q: Why are the elemental strips in parallel and not in series?

Because the voltage $V_0$ is applied across the two flat ends. Every elemental semicircular ring connects these same two ends, meaning they all have the same potential difference across them.

Q: Why do we ignore the magnetic field's role in the transverse voltage?

The problem explicitly states to consider "purely kinetic effects". In reality, the current creates a magnetic field which causes a Hall effect, but we are asked to isolate the effect of the electrons' inertia requiring a centripetal force.

Q: Why is the electric field pointing outward?

Electrons need an inward centripetal force to move in a circle. Since the force $\mathbf{F} = -e\mathbf{E}$ must be inward, and the charge $-e$ is negative, the electric field vector $\mathbf{E}$ must point in the opposite direction, which is radially outward.

Q: Does a higher electric potential mean a higher potential energy for electrons?

No! Because electrons have a negative charge, they have lower potential energy at higher electric potentials. They tend to move towards higher potential regions.

Q: How did we get $v_d \propto I$?

From the standard current equation $I = neA v_d$. For a given cross-section, $n$, $e$, and $A$ are constants, so the drift velocity $v_d$ is directly proportional to the macroscopic current $I$.

Question

Shown in the figure is a semicircular metallic strip that has thickness t and resistivity $\rho$ . Its inner radius is $R_1$ and outer radius is $R_2$. If a voltage $V_0$ is applied between its two ends, a current I flows in it. In addition, it is observed that a transverse voltage $\Delta V$ develops between its inner and outer surfaces due to purely kinetic effects of moving electrons (ignore any role of the magnetic field due to the current). Then (figure is schematic and not drawn to scale)-

Options:

(A) $I = \frac{V_0 t}{\pi \rho} \ln\left(\frac{R_2}{R_1}\right)$

(B) the outer surface is at a higher voltage than the inner surface

(C) the outer surface is at a lower voltage than the inner surface

(D) $\Delta V \propto I^2$

Sigma Percentile · Accepted Answer

### The Geometry of the Problem

When we think of a resistor, we usually picture a straight cylinder. But in this fascinating problem, we are presented with a semicircular metallic strip. It is connected to a battery across its two flat ends, meaning the current doesn't flow in a straight line; it is forced to travel in a curved, azimuthal path.

This curvature is the heart of the problem. It not only changes how we calculate the resistance but also introduces a beautiful microscopic kinetic effect that we rarely consider in standard circuits.

### Slicing the Resistor

To find the total current $I$, we first need to determine the equivalent resistance of this non-standard shape. We cannot simply use the formula $R = \frac{ho L}{A}$ directly because the length of the path and the cross-sectional area are not uniform if we look at it as a single block.

Instead, we must slice the strip into infinitesimal elements. Imagine the strip is made up of many thin, semicircular wires nested inside each other. Let's consider one such elemental wire at a radial distance $x$ from the center, having an infinitesimally small width $dx$.

### The Calculus of Resistance

For this specific elemental wire, the length it covers is a semicircle, so $l = \pi x$. Its cross-sectional area is a small rectangle of height $t$ (the thickness of the strip) and width $dx$, so $A = t \cdot dx$.

The resistance of this single elemental wire is:

$$dR = \frac{ho l}{A} = \frac{ho (\pi x)}{t \, dx}$$

Now, how are all these elemental wires connected? Since the voltage $V_0$ is applied across the two flat ends of the entire strip, every single elemental wire experiences the exact same potential difference $V_0$. This is the classic definition of a **parallel combination**.

To find the equivalent resistance $R_{eq}$ of resistors in parallel, we integrate their conductances (the reciprocals of resistance):

$$\frac{1}{R_{eq}} = \int_{R_1}^{R_2} \frac{1}{dR} = \int_{R_1}^{R_2} \frac{t \, dx}{ho \pi x}$$

Evaluating this integral yields a natural logarithm:

$$\frac{1}{R_{eq}} = \frac{t}{ho \pi} \ln\left(\frac{R_2}{R_1}ight)$$

### Deriving the Current

With the equivalent resistance in hand, finding the total current is a straightforward application of Ohm's Law, $I = \frac{V_0}{R_{eq}}$:

$$I = \frac{V_0 t}{\pi ho} \ln\left(\frac{R_2}{R_1}ight)$$

This perfectly matches option (A).

### The Kinetic Effect: A Deeper Look

Now, let's shift our focus from macroscopic resistance to microscopic kinematics. Imagine a single electron navigating this curved strip. Because the path is semicircular, the electron is undergoing circular motion with a certain drift velocity $v_d$.

Newton's laws tell us that any object moving in a circle requires a **centripetal force** directed towards the center of the circle. For our electron at radius $x$, this required force is:

$$F_c = \frac{m v_d^2}{x}$$

Where does this force come from? As electrons are forced to turn, their inertia makes them "want" to travel in a straight line. Consequently, they drift slightly towards the outer edge of the strip. This microscopic piling up of negative charge on the outer edge leaves a relative positive charge on the inner edge, generating a **transverse electric field** $E$.

### Potential Difference Direction

This newly generated electric field $E$ exerts an electric force $F_e = -eE$ on the electrons. For this force to act as the centripetal force, it must point radially inward (towards the center).

Because the electron carries a negative charge, the electric field vector $\vec{E}$ must point in the exact opposite direction of the force. Therefore, **the electric field points radially outward**.

By definition, electric field lines point from regions of higher potential to regions of lower potential. Since the field points outward, the inner surface must be at a higher potential than the outer surface.

$$	herefore V_{inner} > V_{outer}$$

This confirms that option (C) is correct.

### The Proportionality of Transverse Voltage

Finally, let's determine how the transverse voltage $\Delta V$ scales with the current $I$. By equating the electric force to the required centripetal force:

$$eE = \frac{m v_d^2}{x} \implies E = \frac{m v_d^2}{e x}$$

This shows that the electric field $E$ is directly proportional to the square of the drift velocity ($E \propto v_d^2$).

The transverse voltage $\Delta V$ is simply the integral of this electric field across the width of the strip ($\Delta V = \int E \, dx$). Therefore, $\Delta V$ must also be proportional to $v_d^2$.

Recall the fundamental relation between macroscopic current and microscopic drift velocity: $I = neA v_d$. For a given material and geometry, $I$ is directly proportional to $v_d$.

Substituting this relationship, we arrive at our final elegant conclusion:

$$\Delta V \propto I^2$$

This confirms option (D). The problem beautifully intertwines the calculus of parallel resistors with the mechanics of circular motion applied to charge carriers!

Select Answer:

Visualized Solution

Analyzing the Geometry

Elemental Semicircular Ring

Resistance of the Element $d R$

Parallel Combination of Elements

Calculating Total Current $I$

Kinetic Effect on Electrons

Transverse Electric Field $E$

Direction of Electric Field

Transverse Voltage $Δ V$ Proportionality

Final Conclusion

The Way Forward: Hall Effect

The Sigma Insight: Ohm's Law, Resistance and Electrical Power

The Geometry of the Problem

Slicing the Resistor

The Calculus of Resistance

Deriving the Current

The Kinetic Effect

Potential Difference Direction

The Proportionality of Transverse Voltage

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List-I

List-II

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Select Answer:

Visualized Solution

Analyzing the Geometry

Elemental Semicircular Ring

Resistance of the Element .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dR

Parallel Combination of Elements

Calculating Total Current .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I

Kinetic Effect on Electrons

Transverse Electric Field .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E

Direction of Electric Field

Transverse Voltage .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔV Proportionality

Final Conclusion

The Way Forward: Hall Effect

The Sigma Insight: Ohm's Law, Resistance and Electrical Power

The Geometry of the Problem

Slicing the Resistor

The Calculus of Resistance

Deriving the Current

The Kinetic Effect

Potential Difference Direction

The Proportionality of Transverse Voltage

Similar Questions

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Transverse Voltage $Δ V$ Proportionality

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