Coefficient of mutual inductance between two given square coils is proportional to $(L_{2} >> L_{1})$

\frac{L_{1}}{L_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{L_{1}^{2}}{L_{2}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{L_{1}}{L_{2}^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

L_{1} L_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{L_{1}^{2}}{L_{2}}$
$\begin{matrix} f l u x t h r o u g h c o i l l ϕ = B A N o w B = \frac{μ_{o} i (sin 45^{o} + sin 45^{0})}{4 π (\frac{L_{2}}{2})} \times 4 = \frac{2 μ_{o} i (\sqrt{2}) \times 4}{4 π (L_{2})} = \frac{2 \sqrt{2} μ_{o} i}{π L_{2}} ϕ = \frac{2 \sqrt{2} μ_{o} i}{π L_{2}} \times {L_{1}}^{2} ∴\Rightarrow M = \frac{2 \sqrt{2} μ_{o} {L_{1}}^{2}}{π L_{2}} ∴ M \propto \frac{{L_{1}}^{2}}{L_{2}} H e n c e, o p t i o n B i s c o r r e c t a n s w e r . \end{matrix}$

Suggest Corrections

Similar questions

P

S

L_{1}

L_{2}

M = \sqrt{L_{1} L_{2}}

L_{1}

L_{2}

M

The representation of $l_{1} parallel to l_{2}$ and $l_{3} perpendicular to l_{2}$ is respectively,

l_{1} a n d l_{2}

L_{1}

L_{2}

M

M =

Join BYJU'S Learning Program