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JEE Main Physics Class 12 Moving Charges and Magnetism Part 2 Questions
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Question : 57
Total: 100
There are two infinitely long straight current carrying conductors and they are held at right angles to each other so that their common ends meet at the origin as shown in the figure given below. The ratio of current in both conductors is
1
:
1
. Themagnetic field at point P is
[1 Sep 2021 Shift 2]
μ
0
I
4
π
x
y
[
√
x
2
+
y
2
+
(
x
+
y
)
]
μ
0
I
4
π
x
y
[
√
x
2
+
y
2
−
(
x
+
y
)
]
μ
0
Ixy
4
π
[
√
x
2
+
y
2
−
(
x
+
y
)
]
μ
0
Ixy
4
π
[
√
x
2
+
y
2
+
(
x
+
y
)
]
Validate
Solution:
The diagram is given below,
The magnetic field due to the wire 1,
B
wire
1
=
μ
0
I
4
π
y
(
sin
90
°
+
sin
θ
1
)
μ
0
I
4
π
y
(
1
+
x
√
x
2
+
y
2
)
The magnetic field due to the wire 2,
μ
0
I
4
π
y
(
sin
90
°
+
sin
θ
2
)
B
wire
2
=
μ
0
I
4
π
x
(
1
+
y
√
x
2
+
y
2
)
Thus, the total magnetic field,
B
=
B
1
+
B
2
=
μ
0
I
4
π
(
1
y
+
x
y
√
x
2
+
y
2
+
1
x
+
y
x
√
x
2
+
y
2
)
=
μ
0
I
4
π
x
y
[
√
x
2
+
y
2
+
(
x
+
y
)
]
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