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AIEEE 2008 Solved Paper
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© examsnet.com
Question : 57
Total: 104
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
′
∆
V
′
developed between
′
B
′
and
′
C
′
. The calculation is done in the following steps:
(i) Take current
′
I
′
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
)
at
r
.
(iv) Repeat (i), (ii) and (iii) for current
′
I
′
leaving
′
D
′
and superpose results for
′
A
′
and
′
D
′
.
ρ
I
π
a
−
ρ
I
π
(
a
+
b
)
ρ
I
a
−
ρ
I
(
a
+
b
)
ρ
I
2
π
a
−
ρ
I
2
π
(
a
+
b
)
ρ
I
2
π
(
a
−
b
)
Validate
Solution:
Let
j
be the current density.
Then
j
×
2
π
r
2
=
I
⇒
j
=
I
2
π
r
2
∴
E
=
ρ
j
=
ρ
I
2
π
r
2
Now,
∆
V
B
C
′
=
−
a
∫
a
+
b
→
E
⋅
→
d
r
=
−
a
∫
a
+
b
ρ
I
2
π
r
2
d
r
=
−
ρ
I
2
π
[
−
1
r
]
a
+
b
a
=
ρ
I
2
π
a
−
ρ
I
2
π
(
a
+
b
)
On applying superposition as mentioned we get
∆
V
B
C
=
2
×
∆
V
B
C
=
ρ
I
π
a
−
ρ
I
π
(
a
+
b
)
© examsnet.com
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