Two very tiny wire loops- with areas a-1- and a-2- are situated a displacement r- apartsee figure- Their mutual inductance is given by-

The correct option is A μ_0/4π r^3[3(a⃗_⃗1⃗·r̂)(a⃗_⃗2⃗·r̂)+a⃗_⃗1⃗·a⃗_⃗2⃗] 2 magnetic field due to loop at a distance r is #160; B _ 1 =μ #16 ...

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Standard XII

Mathematics

Property 2

Two very tiny...

Question

Two very tiny wire loops, with areas $_{1}$ and $_{2}$ are situated a displacement $\to r$ apart(see figure). Their mutual inductance is given by.

\frac{μ_{0}}{4 π r^{3}} [3 (_{1} \cdot^r) (_{2} \cdot^r) +_{1} \cdot_{2}]

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\frac{3 μ_{0}}{4 π r^{3}} [3 (_{1} \cdot^r) (_{2} \cdot^r) +_{1} \cdot_{2}]

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\frac{3 μ_{0}}{4 π r^{3}} [3 (_{1} \cdot^r) (_{2} \cdot^r -_{1} \cdot_{2}]

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\frac{μ_{0}}{4 π r^{3}} [3 (_{1} \cdot^r) (_{2} \cdot^r) -_{1} \cdot_{2}]

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Solution

The correct option is A $\frac{μ_{0}}{4 π r^{3}} [3 (_{1} \cdot^r) (_{2} \cdot^r) +_{1} \cdot_{2}]$
2 magnetic field due to loop $⊥$ at a distance r is
$B_{1} = \frac{μ_{0} I_{1}}{4 π r^{3}} [3 (_{1}^r) r - a]$ if current $I_{1}$ is flowing through Loop 1
Flux through Loop 2
$ϕ_{2} = B_{1} . a_{2} = \frac{μ_{0}}{4 π r^{3}} I_{1} (3 (_{1} .^r) (_{2} .^r) -_{1} ._{2})$
Mutual inductance $M = \frac{ϕ_{2}}{I_{1}}$
$= \frac{μ_{0}}{4 π r^{3}} [3 (_{1} .^r) (_{2} .^r) +_{1} ._{2}]$

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Two very tiny wire loops, with areas →a1 and →a2 are situated a displacement →r apart(see figure). Their mutual inductance is given by.

Two very tiny wire loops, with areas $_{1}$ and $_{2}$ are situated a displacement $\to r$ apart(see figure). Their mutual inductance is given by.