What is the magnitude of the net force on the first wire in (Figure 1)?
Answer:
You can see from the diagram that, wire 1 has an attracting force from wire 3 at 0.04 m distance and a repelling force from wire 2 at 0.02 m distance.
Expression of the force is –
F = µI1I2*L/(2πd)
When d = 0.02 m,
F = (4pi10^-710100.50) / (2pi*0.02) = 0.0005 N
And when d = 0.04 m,
F = (4pi10^-710100.50) / (2pi*0.04) = 0.00025 N
(a) So, the net force on wire 1 = 0.005 – 0.00025 = 0.00025 N repelling, or upward.
(c) By symmetry the net force on wire 3 is the same means, 0.00025 N, but downward direction.
(b) Wire 2 has equal repelling forces from wires 1 and 3 so net force = 0.
What is the magnitude of the net force on the first wire in (Figure 1)?
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