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Current Electricity question

2008 · Shift 0 · Q69
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Current Electricity question

2008 · Shift 0 · Q69

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 309 English For current entering at A,A,A, the electric field at a distance ′r′'r'′r′ from AAA is
  1. A
    ρI8πr2{{\rho I} \over {8\pi {r^2}}}8πr2ρI​
  2. B
    ρIr2{{\rho I} \over {{r^2}}}r2ρI​
  3. C
    ρI2πr2{{\rho I} \over {2\pi {r^2}}}2πr2ρI​
  4. D
    ρI4πr2{{\rho I} \over {4\pi {r^2}}}4πr2ρI​
View written solutionFree

Correct answer: C

  1. Current spreading from point AAA

    Since the current enters the conducting block from a point on the surface, it spreads into the material through a hemispherical surface of radius rrr.

    The area of a hemisphere of radius rrr is A=2πr2.A = 2\pi r^2.A=2πr2.

  2. Current density at distance rrr

    If total current is III, then the current density at radius rrr is j(r)=I2πr2.j(r) = \frac{I}{2\pi r^2}.j(r)=2πr2I​.

  3. Using Ohm's law in microscopic form

    We are given E⃗=ρj⃗.\vec E = \rho \vec j.E=ρj​.

    Therefore, the magnitude of electric field is E(r)=ρj(r)=ρ⋅I2πr2.E(r) = \rho j(r) = \rho \cdot \frac{I}{2\pi r^2}.E(r)=ρj(r)=ρ⋅2πr2I​.

    Hence, E(r)=ρI2πr2.E(r) = \frac{\rho I}{2\pi r^2}.E(r)=2πr2ρI​.

  4. Matching with options

    This corresponds to Option C.

Comparison with stored answer: Stored correct answer is C, which matches our derived result.

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