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Current Electricity question

2008 · Shift 0 · Q67
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Current Electricity question

2008 · Shift 0 · Q67

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
Consider a block of conducting material of resistivity ′ρ′'\rho '′ρ′ shown in the figure. Current ′I′'I'′I′ enters at ′A′'A'′A′ and leaves from ′D′'D'′D′. We apply superposition principle to find voltage ′ΔV′'\Delta V'′ΔV′ developed between ′B′'B'′B′ and ′C′'C'′C′. The calculation is done in the following steps: (i) Take current ′I′'I'′I′ entering from ′A′'A'′A′ and assume it to spread over a hemispherical surface in the block. (ii) Calculate field E(r)E(r)E(r) at distance ′r′'r'′r′ from A by using Ohm's law E=ρj,E = \rho j,E=ρj, where jjj is the current per unit area at ′r′'r'′r′. (iii) From the ′r′'r'′r′ dependence of E(r)E(r)E(r), obtain the potential V(r)V(r)V(r) at rrr. (iv) Repeat (i), (ii) and (iii) for current ′I′'I'′I′ leaving ′D′'D'′D′ and superpose results for ′A′'A'′A′ and ′D′.'D'.′D′.AIEEE 2008 Physics - Current Electricity Question 310 EnglishΔV\Delta VΔV measured between BBB and CCC is
  1. A
    ρIπa−ρIπ(a+b){{\rho I} \over {\pi a}} - {{\rho I} \over {\pi \left( {a + b} \right)}}πaρI​−π(a+b)ρI​
  2. B
    ρIa−ρI(a+b){{\rho I} \over a} - {{\rho I} \over {\left( {a + b} \right)}}aρI​−(a+b)ρI​
  3. C
    ρI2πa−ρI2π(a+b){{\rho I} \over {2\pi a}} - {{\rho I} \over {2\pi \left( {a + b} \right)}}2πaρI​−2π(a+b)ρI​
  4. D
    ρI2π(a−b){{\rho I} \over {2\pi \left( {a - b} \right)}}2π(a−b)ρI​
View written solutionFree

Correct answer: A

  1. Current spreading from a point contact on the surface

Since current enters the conducting block at point AAA on the surface, it spreads into the material through hemispherical surfaces.

At a distance rrr from AAA, the area of a hemisphere is A(r)=2πr2.A(r)=2\pi r^2.A(r)=2πr2. So the current density is j(r)=I2πr2.j(r)=\frac{I}{2\pi r^2}.j(r)=2πr2I​.

Using Ohm’s law in microscopic form, E⃗=ρj⃗,\vec E = \rho \vec j,E=ρj​, we get the magnitude E(r)=ρj(r)=ρI2πr2.E(r)=\rho j(r)=\frac{\rho I}{2\pi r^2}.E(r)=ρj(r)=2πr2ρI​.

  1. Potential due to current entering at AAA

Potential difference between two points at distances r1r_1r1​ and r2r_2r2​ from AAA is

=\int_{r_1}^{r_2} \frac{\rho I}{2\pi r^2}\,dr.$$ Thus, $$V(r)=\frac{\rho I}{2\pi r}+\text{constant}.$$ So the contribution of source at $A$ is $$V_A(r)=\frac{\rho I}{2\pi r}.$$ --- 3. **Potential due to current leaving at $D$** If current leaves from $D$, that acts like a sink. Its contribution to potential at distance $r$ from $D$ is $$V_D(r)=-\frac{\rho I}{2\pi r}.$$ Hence total potential at any point is $$V=\frac{\rho I}{2\pi (\text{distance from }A)}-\frac{\rho I}{2\pi (\text{distance from }D)}.$$ --- 4. **Evaluate potentials at $B$ and $C$** From the figure geometry (standard arrangement along the surface): - $AB=a$ - $BC=b$ - therefore $AC=a+b$ - also, by symmetry with terminals at the ends, distance of $B$ from $D$ is $a+b$ - distance of $C$ from $D$ is $a$ So, ### Potential at $B$ $$V_B=\frac{\rho I}{2\pi a}-\frac{\rho I}{2\pi (a+b)}.$$ ### Potential at $C$ $$V_C=\frac{\rho I}{2\pi (a+b)}-\frac{\rho I}{2\pi a}.$$ --- 5. **Voltage between $B$ and $C$** Therefore, $$\Delta V = V_B-V_C$$ $$=\left(\frac{\rho I}{2\pi a}-\frac{\rho I}{2\pi (a+b)}\right)-\left(\frac{\rho I}{2\pi (a+b)}-\frac{\rho I}{2\pi a}\right).$$ Simplifying, $$\Delta V=\frac{\rho I}{\pi a}-\frac{\rho I}{\pi (a+b)}.$$ This matches **Option A**. --- 6. **Comparison with stored answer** Derived answer: **A** Stored correct answer: **A** So they agree.
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