Two straight, parallel, current-carrying wires separated by a distance of 5cm contain the same 1000 A current, flowing in similar directions. Determine the force per unit length exerted by the two wires.
We have given that,
I1=I2=1000AI_1=I_2 = 1000AI1=I2=1000A
r=5cm=0.05mr = 5cm = 0.05mr=5cm=0.05m
We know the formula,
Fl=μ0I1I22πr\dfrac{F}{l} = \dfrac{\mu_0I_1I_2}{2\pi r}lF=2πrμ0I1I2 =4π×10−7×1000×10002×3.14×0.05=4Nm−1= \dfrac{4\pi \times 10^{-7}\times1000 \times 1000 }{2 \times 3.14 \times0.05} = 4Nm^{-1}=2×3.14×0.054π×10−7×1000×1000=4Nm−1
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