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UPSEE Solved Model Paper 3
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© examsnet.com
Question : 5
Total: 150
Let BP and BQ be the magnetic field produced by the wire P and Q which are placed symmetrically in a rectangular loop ABCD as shown in figure. Current in wire P is I directed inward and in Q is 2 I directed outwards. If
∫
A
B
B
Q
.
d
l
=
+
2
µ
0
T
−
m
,
∫
D
A
B
P
.
d
l
=
−
2
µ
0
T
−
m
and
∫
A
B
B
P
.
d
l
=
−
µ
0
T
−
m
.
the value of I will be
8A
4A
5A
6A
Validate
Solution:
From diagram,
∫
A
B
B
P
d
I
+
∫
B
C
B
P
d
i
+
∫
C
D
B
P
d
I
+
∫
D
A
B
P
d
I
+
∫
A
B
B
Q
d
I
+
∫
B
C
B
Q
d
I
+
∫
C
D
B
Q
d
I
+
∫
D
A
B
Q
d
I
=
µ
0
(
2
I
−
I
)
(
−
µ
0
−
2
µ
0
−
µ
0
−
2
µ
0
)
+
(
2
µ
0
+
4
µ
0
+
2
µ
0
+
4
µ
0
)
=
µ
0
I
⇒ I = 12 - 6 = 6A
© examsnet.com
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