Shown in the figure is a semicircular metallic strip that has thickness t and resistivity \rho . Its inner radius is R1 and outer radius is R2. If a voltage V0 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)

I = V_0t \over \pi \rho \ln ( R_2 \over R_1 ), the outer surface is at a lower voltage than the inner surface, \Delta V \propto I2

I = V_0t \over \pi \rho \ln ( R_2 \over R_1 )

the outer surface is at a higher voltage than the inner surface

the outer surface is at a lower voltage than the inner surface

Current Electricity PYQ — JEE Advanced 2020 · Shift 1 · Q46 with Solution

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Correct answer: A, C, D

Current through the semicircular strip

The current flows along the semicircular direction, i.e. along arcs of radius $r$ .

For a thin annular element between radii $r$ and $r+dr$ :

length of current path $= \pi r$
cross-sectional area perpendicular to current $= t\,dr$

So its resistance is $dR=\frac{\rho (\pi r)}{t\,dr}$

But these annular strips are connected in parallel between the two ends of the semicircle, because each strip spans the same potential difference $V_0$ .

Hence it is better to add conductances: $dG=\frac{1}{dR}=\frac{t}{\rho \pi}\frac{dr}{r}$

Integrating from $R_1$ to $R_2$ ,

=\frac{t}{\rho \pi}\ln\left(\frac{R_2}{R_1}\right)$$ Therefore, $$I=GV_0=\frac{V_0 t}{\pi \rho}\ln\left(\frac{R_2}{R_1}\right)$$ So **Option A is correct**. --- 2. **Drift speed at radius $r$** Let the current through the annular strip of width $dr$ be $dI$. Since all strips have the same potential difference, $$dI = V_0\, dG = V_0\frac{t}{\rho \pi}\frac{dr}{r}$$ Current density along the strip at radius $r$ is $$J(r)=\frac{dI}{t\,dr}=\frac{V_0}{\rho \pi r}$$ Using $$J = n e v_d$$ (in magnitude), we get $$v_d(r)=\frac{J(r)}{ne}=\frac{V_0}{\rho \pi ne}\frac{1}{r}$$ Thus, $$v_d \propto \frac{1}{r}$$ So electrons move faster near the inner radius and slower near the outer radius. --- 3. **Transverse voltage due to purely kinetic effect** Because electrons are moving in a curved path, they require centripetal acceleration. This can be provided by a radial electric field. For an electron moving along a circular arc, $$\frac{mv_d^2}{r} = e E_r$$ Here $E_r$ must be directed inward for the electron force to be inward. Since electron charge is negative, the electric field itself must be directed outward. If electric field is outward, potential decreases outward because $$\vec E = -\nabla V$$ Hence the **outer surface is at lower potential than the inner surface**. So: - **B is false** - **C is true** --- 4. **Dependence of transverse voltage on current** From above, $$E_r = \frac{m v_d^2}{e r}$$ Since $v_d \propto I$ (because $J \propto I$), we get $$E_r \propto I^2$$ More explicitly, since $v_d(r) \propto 1/r$, $$E_r(r) \propto \frac{1}{r^3}$$ Then transverse voltage between inner and outer radii is $$\Delta V = \int_{R_1}^{R_2} E_r\,dr$$ The radial dependence integrates to a geometric factor involving $R_1,R_2$, but the current dependence remains $$\Delta V \propto I^2$$ So **Option D is correct**. --- 5. **Final evaluation of options** - **A:** Correct - **B:** Incorrect - **C:** Correct - **D:** Correct Therefore the correct options are $$\boxed{A,\ C,\ D}$$ --- 6. **Comparison with stored correct answer** Stored correct answer: **A, C, D** My derived answer: **A, C, D** They agree.

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View written solutionFree

Correct answer: A, C, D

Current through the semicircular strip

The current flows along the semicircular direction, i.e. along arcs of radius $r$ .

For a thin annular element between radii $r$ and $r+dr$ :

length of current path $= \pi r$
cross-sectional area perpendicular to current $= t\,dr$

So its resistance is $dR=\frac{\rho (\pi r)}{t\,dr}$

But these annular strips are connected in parallel between the two ends of the semicircle, because each strip spans the same potential difference $V_0$ .

Hence it is better to add conductances: $dG=\frac{1}{dR}=\frac{t}{\rho \pi}\frac{dr}{r}$

Integrating from $R_1$ to $R_2$ ,

=\frac{t}{\rho \pi}\ln\left(\frac{R_2}{R_1}\right)$$ Therefore, $$I=GV_0=\frac{V_0 t}{\pi \rho}\ln\left(\frac{R_2}{R_1}\right)$$ So **Option A is correct**. --- 2. **Drift speed at radius $r$** Let the current through the annular strip of width $dr$ be $dI$. Since all strips have the same potential difference, $$dI = V_0\, dG = V_0\frac{t}{\rho \pi}\frac{dr}{r}$$ Current density along the strip at radius $r$ is $$J(r)=\frac{dI}{t\,dr}=\frac{V_0}{\rho \pi r}$$ Using $$J = n e v_d$$ (in magnitude), we get $$v_d(r)=\frac{J(r)}{ne}=\frac{V_0}{\rho \pi ne}\frac{1}{r}$$ Thus, $$v_d \propto \frac{1}{r}$$ So electrons move faster near the inner radius and slower near the outer radius. --- 3. **Transverse voltage due to purely kinetic effect** Because electrons are moving in a curved path, they require centripetal acceleration. This can be provided by a radial electric field. For an electron moving along a circular arc, $$\frac{mv_d^2}{r} = e E_r$$ Here $E_r$ must be directed inward for the electron force to be inward. Since electron charge is negative, the electric field itself must be directed outward. If electric field is outward, potential decreases outward because $$\vec E = -\nabla V$$ Hence the **outer surface is at lower potential than the inner surface**. So: - **B is false** - **C is true** --- 4. **Dependence of transverse voltage on current** From above, $$E_r = \frac{m v_d^2}{e r}$$ Since $v_d \propto I$ (because $J \propto I$), we get $$E_r \propto I^2$$ More explicitly, since $v_d(r) \propto 1/r$, $$E_r(r) \propto \frac{1}{r^3}$$ Then transverse voltage between inner and outer radii is $$\Delta V = \int_{R_1}^{R_2} E_r\,dr$$ The radial dependence integrates to a geometric factor involving $R_1,R_2$, but the current dependence remains $$\Delta V \propto I^2$$ So **Option D is correct**. --- 5. **Final evaluation of options** - **A:** Correct - **B:** Incorrect - **C:** Correct - **D:** Correct Therefore the correct options are $$\boxed{A,\ C,\ D}$$ --- 6. **Comparison with stored correct answer** Stored correct answer: **A, C, D** My derived answer: **A, C, D** They agree.

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