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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= [( 0 I )/ ] log e (d-a)/d b. L= [( 0 I )/ ] log e d/a c. L= [( 0 I )/ ] log e a/d d. L= [( 0 )/ ] Id/a


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


Grade:10

1 Answers

ROSHAN MUJEEB
askIITians Faculty 829 Points
2 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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