Answer:
Magnetic field at O due to wire 1, B
1
=
2πd
μ
o
I
into the paper
where d=0.2 cm=0.002 m
∴ B
2π(0.002)
(8)
Magnetic field at O due to wire 2, B
2
out of the paper
(12)
Thus net magnetic field at O B
O
=B
−B
(12−8)
(out of paper)
(4π×10
−7
)(4)
=4.0×10
−3
T
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Answers & Comments
Answer:
Magnetic field at O due to wire 1, B
1
=
2πd
μ
o
I
1
into the paper
where d=0.2 cm=0.002 m
∴ B
1
=
2π(0.002)
μ
o
(8)
into the paper
Magnetic field at O due to wire 2, B
2
=
2πd
μ
o
I
2
out of the paper
∴ B
2
=
2π(0.002)
μ
o
(12)
out of the paper
Thus net magnetic field at O B
O
=B
2
−B
1
=
2π(0.002)
μ
o
(12−8)
(out of paper)
∴ B
O
=
2π(0.002)
(4π×10
−7
)(4)
=4.0×10
−3
T