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Updated on : 2022-09-05

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If $(t)$ is instantaneous current then,

$I(t)=R1 dtdϕ ....(I)$

If $Q$ is charge passing in time $t$

$∴I(t)=dtdQ ....(II)$

From (I) and (II)

$dtdQ =R1 ⋅dtdϕ $

Or $dQ=R1 .dϕ....(III)$

Integrating both sides,

$∫_{Q_{1}}dQ=R1 ∫_{ϕ_{1}}dϕ$

$Q_{2}(t)−Q_{1}(t)=R1 [ϕ_{2}(t)−ϕ_{1}(t)]$

For magnetic flux in rectangle:

Magnetic flux due to current carrying conductor at a distance $x$'

$Q(t)=2πx_{′}μ_{0}I(t) $

If length of strip is $L_{1}$ so total flux on strip of length $L_{1}$ at distance $x_{′}$ is

$Q(t)=2πx_{′}μ_{0}I(t) L_{1}$

$X_{′}$ varies from $x$ to $(x+L_{2})$ so total flux in strip

$ϕ(t)=2πμ_{0} ∫_{x}x_{′}dx I(t)=2πμ_{0}L_{2} .I(t_{1})g_{e}x(L_{2}+x) $

The magnitude of charge is given on length $L_{1}$

$∫_{Q_{1}}dQ=R1 ∫dϕ$ [from (III)]

$∫_{0}dQ=R1 ⋅2πμ_{0}L_{1} g_{e}(xL_{2}+x )∫_{0}I(t_{1})$

$Q=R2πμ_{0}L_{1} g_{e}(xL_{2}+x )(I−0)=2πRμ_{0}L_{1}L_{1} g(xL_{2}+x )$.

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