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JEE Main 9 Apr 2019 Paper 1 Solved Paper
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© examsnet.com
Question : 24
Total: 90
A rigid square loop of side ‘a’ and carrying current
I
2
is lying on a horizontal surface near a long current
I
1
carrying wire in the same plane as shown in figure.
The net force on the loop due to the wire will be:
Attractive and equal to
µ
0
I
1
I
2
3
π
Zero
Repulsive and equal to 0
µ
0
I
1
I
2
4
π
Repulsive and equal to
µ
0
I
1
I
2
2
π
Validate
Solution:
–
–
–
–
F
A
B
+
–
–
–
–
F
D
C
=
0
F
A
D
=
B
I
2
a
=
µ
0
I
2
π
(
α
)
I
2
(
α
)
(right wards)
=
µ
I
1
I
2
2
π
F
B
C
=
B
,
I
2
a
=
µ
0
I
1
I
2
4
π
(up + wards)
F
n
e
t
=
µ
0
I
1
I
2
2
π
−
µ
0
I
1
I
2
4
π
=
µ
0
I
1
I
2
4
pi
(right wards ev (repulsive)
© examsnet.com
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