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Intensity Amplitude Relationship

The mutual in...

Question

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}}

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\frac{μ_{0} π r_{1}^{2}}{2 r_{2}}

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\frac{μ_{0} π (r_{1} + r_{2})^{2}}{2 r_{1}}

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\frac{μ_{0} π (r_{1} + r_{2})^{2}}{2 r_{2}}

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Solution

The correct option is A $\frac{μ_{0} π r_{2}^{2}}{2 r_{1}}$
We have,
$\begin{matrix} ϕ_{21} = \frac{μ_{0} i}{2 r_{1}} \times π r_{2}^{2} ∴ μ = \frac{μ_{0} i π r_{2}^{2}}{2 r_{1}} \end{matrix}$
$∴$ Option $A$ is correct.

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The mutual inductance between two planar concentric rings of radii $r_{1} a n d r_{2} (w i t h r_{1} >> r_{2})$ placed in air is given by

If the sum of the areas of two circles with radii R₁ and R₂ is equal to the area of a circle of radius R, then

If the sum of the areas of two circles with radii $R_{1}$ and $R_{2}$ is equal to the area of a circle of radius R then

(a) $R_{1} + R_{2} = R$

(b) $R_{1} + R_{2} > R$

(d) $R_{1}^{2} + R_{2}^{2} = R^{2}$

R_{1}

R_{2}

R_{1} + R_{2} = R

R_{1}^{2} + R_{2}^{2} = R^{2}

R_{1} + R_{2} < R

R_{1}^{2} + R_{2}^{2} < R_{2}

r_{1}

r_{2}

r

r = r_{1} + r_{2}

{r_{1}}^{2} + {r_{2}}^{2} = r^{2}

r_{1} + r_{2} < r

{r_{1}}^{2} + {r_{2}}^{2} < r^{2}

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