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JEE Main 2008 (AIEEE) · Ohm's Law, Resistivity & Temperature

Ohm's Law, Resistivity & Temperature question from JEE Main 2008 (AIEEE)

Q63JEE Main 2008 · AIEEEDifficult

Ohm's Law, Resistivity & Temperature

Consider a block of conducting material of resistivity ’ρ’\text{'}\rho\text{'} shown in the figure. Current ’I’\text{'}I\text{'} enters at ’A’\text{'}A\text{'} and leaves from ’D’\text{'}D\text{'}. We apply superposition principle to find voltage ’ΔV’\text{'}\Delta V\text{'} developed between ’B’\text{'}B\text{'} and ’C’\text{'}C\text{'}. The calculation is done in the following steps:

(i) Take current ’I’\text{'}I\text{'} entering from ’A’\text{'}A\text{'} and assume it to spread over a hemispherical surface in the block.

(ii) Calculate field E(r)E(r) at distance ’r’\text{'}r\text{'} from A by using Ohm's law E=ρj,E = \rho j, where jj is the current per unit area at ’r’\text{'}r\text{'}.

(iii) From the ’r’\text{'}r\text{'} dependence of E(r)E(r), obtain the potential V(r)V(r) at rr.

(iv) Repeat (i), (ii) and (iii) for current ’I’\text{'}I\text{'} leaving ’D’\text{'}D\text{'} and superpose results for ’A’\text{'}A\text{'} and ’D’.\text{'}D\text{'}.

Figure

ΔV\Delta V measured between BB and CC is

Show answer and solution

Answer: (a)

A (+I)BCD (−I)abaV(r) = ±ρI / (2πr) from each contact
  1. Current II at A spreads over a hemisphere: j=I2πr2j = \dfrac{I}{2\pi r^2}, E=ρjE = \rho j, so V(r)=ρI2πrV(r) = \dfrac{\rho I}{2\pi r}.
  2. Due to A: VB−VC=ρI2π(1a−1a+b)V_B - V_C = \dfrac{\rho I}{2\pi}\left(\dfrac1a - \dfrac1{a+b}\right).
  3. Due to D (current leaving, V=−ρI2πrV = -\dfrac{\rho I}{2\pi r}; B is at a+ba+b, C at aa from D): VB−VC=ρI2π(1a−1a+b)V_B - V_C = \dfrac{\rho I}{2\pi}\left(\dfrac1a - \dfrac1{a+b}\right).
  4. Adding: ΔV=ρIπa−ρIπ(a+b)\Delta V = \dfrac{\rho I}{\pi a} - \dfrac{\rho I}{\pi(a+b)}. The answer is (a).
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