JEE Main · 2008 · 27 Apr 2008 · Offline · Current Electricity

Passage

Consider a block of conducting material of resistivity '

ρ

' shown in the figure. Current ' I ' enters at 'A' and leaves from ' D '. We apply superposition principle to find voltage '

Δ V

' developed between ' B ' and ' C '. The calculation is done in the following steps:
(i) Take current ' l ' entering from ' A ' and assume it to spread over a hemispherical surface in the block.
(ii) Calculate field

E (r)

at distance '

r

' from A by using Ohm's law

E = ρ j

, where

j

is the current per unit area at ' r '.
(iii) From the '

r

' dependence of

E (r)

, obtain the potential

V (r)

r

.
(iv) Repeat (i), (ii) and (iii) for current ' I ' leaving ' D ' and superpose results for ' A ' and ' D '.

For current entering at A , the electric field at a distance ' r ' from A is

(A) $\frac{ρl}{8 π r ^{2}}$
(B) $\frac{ρl}{r ^{2}}$
(C) $\frac{ρl}{2 π r ^{2}}$ ✓ Correct
(D) $\frac{ρl}{4 π r ^{2}}$

Answer: (C)

Solution

The current

I

entering at

A

is taken to spread uniformly over a hemispherical surface, which at radius

r

has area

2 π r^{2}

. The current per unit area is therefore

j = \frac{I}{2 π r ^{2}}

, and Ohm's law in its local form

E = ρ j

turns this into a field

E = \frac{ρ I}{2 π r ^{2}}

at distance

r

from

A

JEE Main · 2008 · 27 Apr 2008 · Offline · Single correct option

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JEE Main · 2008 · 27 Apr 2008 · Offline · Current Electricity

Solution

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