$\Delta \mathrm{V}$ measured between B and C is

(C) $\frac{\rho \mathrm{l}}{2 \pi \mathrm{a}}-\frac{\rho \mathrm{l}}{2 \pi(\mathrm{a}+\mathrm{b})}$

(A) $\frac{\rho \mathrm{l}}{\pi \mathrm{a}}-\frac{\rho \mathrm{l}}{\pi(\mathrm{a}+\mathrm{b})}$

(B) $\frac{\rho l}{a}-\frac{\rho l}{(a+b)}$

(D) $\frac{\rho l}{2 \pi(a-b)}$

Current Electricity — JEE Main · 2008 · 27 Apr 2008 · Offline Question with Solution

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 '.

Δ V

measured between B and C is

(A) $\frac{ρ l}{π a} - \frac{ρ l}{π ( a + b )}$
(B) $\frac{ρl}{a} - \frac{ρl}{( a + b )}$
(C) $\frac{ρ l}{2 π a} - \frac{ρ l}{2 π ( a + b )}$ ✓ Correct
(D) $\frac{ρl}{2 π ( a - b )}$

Answer: (C)

Solution

Following the prescribed scheme, the current entering at

A

spreads over a hemisphere, so at radius

r

the field is

E = ρ j = \frac{ρ I}{2 π r ^{2}}

and the potential it sets up is

V (r) = \frac{ρ I}{2 π r}

. Point

B

lies at distance

a

from

A

and point

C

at distance

a + b

, so the potential difference developed between them is

V_{B} - V_{C} = \frac{ρ I}{2 π} (\frac{1}{a} - \frac{1}{a + b})

, which is the voltage measured between

B

and

C

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

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

Solution

More JEE Main questions on Current Electricity