Three long straight and parallel wires
carrying currents are arranged as shown in
fig. The force experienced by 10 cm length
of wire Qis
R Q
P
2cm
10cm
20A 10A
30A
Answers
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Answer:
Magnetic field produced by wire R at Q: B
1
=
2πr
1
μ
0
i
1
=
2π×(2×10
−2
)
μ
0
×20
Magnetic field produced by wire P at Q: B
2
=
2πr
2
μ
0
i
2
=
2π×(10×10
−2
)
μ
0
×30
B
net
=B
1
−B
2
=
2π×10
−2
μ
0
×7
Force on 10 cm of wire at Q: F
Q
=i
Q
lB
net
F=
2π×10×10
−2
4π×10
−7
×20×10×10×10
−2
=
2π×10
−2
4π×10
−7
×10×10
−2
[100−30]
=20×10
−7
×70=1400×10
−7
=1.4×10
−4
N toward right.
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