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JEE Advanced 2023 Paper 2 Solutions
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© examsnet.com
Question : 9
Total: 52
A rectangular conducting loop of length
4
cm
and width
2
cm
is in the
x
y
-plane, as shown in the figure. It is being moved away from a thin and long conducting wire along the direction
√
3
2
^
x
+
1
2
^
y
with a constant speed
v
. The wire is carrying a steady current
I
=
10
A
in the positive
x
-direction. A current of
10
µ
A
flows through the loop when it is at a distance
d
=
4
cm
from the wire. If the resistance of the loop is
0.1
Ω
, then the value of
v
is
ms
−
1
.
[Given: The permeability of free space
µ
0
=
4
π
×
10
−
7
NA
−
2
]
Your Answer:
Validate
Solution:
The two sides perpendicular to the wire would contribute net zero emf. For parallel sides:
→
E
=
→
B
×
→
V
=
µ
0
l
2
π
x
×
V
⇒
Net emf
=
(
E
1
cos
60
∘
−
E
2
cos
60
∘
)
×
width
=
1
2
×
2
100
×
µ
0
∕
v
2
π
[
1
4
100
−
1
8
100
]
=
1
100
×
10
−
7
×
2
×
10
×
v
×
100
×
1
8
=
2.5
v
×
10
−
7
=
i
×
R
⇒
v
=
10
×
10
−
6
×
0.1
2.5
×
10
−
7
=
4
m
∕
s
© examsnet.com
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