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Two long parallel wires whose centres are a distance 'd' apart carry equal currents in opposite directions.If the flux within wires is neglected, the inductance of such a wire of length 'l' and radius 'a' will be
a. L= [(0I )/] loge(d-a)/d b. L= [(0I )/] loged/a
c. L= [(0I )/] loge a/d d. L= [(0)/] Id/a

Aditi Chauhan , 11 Years ago
Grade 10
anser 1 Answers
ROSHAN MUJEEB

Last Activity: 5 Years ago

Since the wires are infinite, so the system of these two wires can be considered as a closed rectangle of infinite length and breadth equal to d. flux through the strip:
ϕ=∫d−aaμ0I2πr(ldr)=μ0Il2πIn(d−aa)ϕ=∫ad-aμ0I2πr(ldr)=μ0Il2πIn(d-aa)The other wire produces the same result, so the total flux through the dotted rectangle is
ϕ→tal=μ0lIπIn(d−aa)ϕ→tal=μ0lIπIn(d-aa)
The total inducatance of lengthll
L=ϕ→talI=μ0lπIn(d−aa)L=ϕ→talI=μ0lπIn(d-aa)
Inductance per unit length=Ll=μ0πIn(d−aa)
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