The mutual inductance between two planar concentric rings of radii r1 and r2 (with r1r2 ) placed in air is given by

Question

The mutual inductance between two planar concentric rings of radii r1 and r2 (with r1>r2 ) placed in air is given by (A)  (μ 0π r22/2r1)  (B)  (μ 0π r12/2r2)  (C)  (μ 0π (r1+r2)/2r1)  (D)  (μ 0π (r1+r2)2/2r2)

Tardigrade Admin · Accepted Answer

Magnetic field due to the larger coil at its centre is B=(μ 0I/2r1) where I is the current in the larger coil. Flux through the inner coil is φ =B× π r22=(μ 0I/2r1)× π r22 But φ =ML Therefore, M=(μ 0π r22/2r1)

Q. The mutual inductance between two planar concentric rings of radii $r_{1}$ and $r_{2}$ (with $r_{1} > r_{2}$ ) placed in air is given by

$\frac{μ _{0} π r _{2}^{2}}{2 r _{1}}$

$\frac{μ _{0} π r _{1}^{2}}{2 r _{2}}$

$\frac{μ _{0} π ( r _{1} + r _{2} )}{2 r _{1}}$

$\frac{μ _{0} π ( r _{1} + r _{2} ) ^{2}}{2 r _{2}}$

Q. The mutual inductance between two planar concentric rings of radii r1​ and r2​ (with r1​>r2​ ) placed in air is given by

2r1​μ0​πr22​​

2r2​μ0​πr12​​

2r1​μ0​π(r1​+r2​)​

2r2​μ0​π(r1​+r2​)2​

Magnetic field due to the larger coil at its centre is B=2r1​μ0​I​ where I is the current in the larger coil. Flux through the inner coil is ϕ=B×πr22​=2r1​μ0​I​×πr22​ But ϕ=ML Therefore, M=2r1​μ0​πr22​​

Q. The mutual inductance between two planar concentric rings of radii $r_{1}$ and $r_{2}$ (with $r_{1} > r_{2}$ ) placed in air is given by

$\frac{μ _{0} π r _{2}^{2}}{2 r _{1}}$

$\frac{μ _{0} π r _{1}^{2}}{2 r _{2}}$

$\frac{μ _{0} π ( r _{1} + r _{2} )}{2 r _{1}}$

$\frac{μ _{0} π ( r _{1} + r _{2} ) ^{2}}{2 r _{2}}$