Two conducting circular loops of radii R1 and R2 are placed in the same plane with their centres coinciding. If R1R2, the mutual inductance M between them will be directly proportional to:

Question

Two conducting circular loops of radii R1 and R2 are placed in the same plane with their centres coinciding. If R1>>R2, the mutual inductance M between them will be directly proportional to: (A) ( R 1/ R 2) (B) (R2/R1) (C) ( R 12/ R 2) (D) ( R 22/ R 1)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/2a34f4d2fd705596d92db63df6eb674e-.png" /> Magnetic field due to R1 at the center B=(μ0 i/2 R1) Flux linked with R2 φ= BA 2=(μ0 i /2 R 1) × π R 22= Mi M=(μ0 π R22/2 R1) M propto (R22/R1)

Q. Two conducting circular loops of radii $R_{1}$ and $R_{2}$ are placed in the same plane with their centres coinciding. If $R_{1} >> R_{2}$ , the mutual inductance $M$ between them will be directly proportional to:

$\frac{R _{1}}{R _{2}}$

$\frac{R _{2}}{R _{1}}$

$\frac{R _{1}^{2}}{R _{2}}$

$\frac{R _{2}^{2}}{R _{1}}$

Magnetic field due to $R_{1}$ at the center
$B = \frac{μ _{0} i}{2 R _{1}}$
Flux linked with $R_{2}$
$ϕ = B A_{2} = \frac{μ _{0} i}{2 R _{1}} \times π R_{2}^{2} = M i$
$M = \frac{μ _{0} π R _{2}^{2}}{2 R _{1}}$
$M \propto \frac{R _{2}^{2}}{R _{1}}$

Q. Two conducting circular loops of radii R1​ and R2​ are placed in the same plane with their centres coinciding. If R1​>>R2​, the mutual inductance M between them will be directly proportional to:

R2​R1​​

R1​R2​​

R2​R12​​

R1​R22​​