Question

Two planar concentric rings of metal wire having radii r and r2 (r₁ > r2) are placed in air. The current I is flowing through the coil of larger radius. The mutual inductance between the coils is given by ( μο = permeability of free space) ______.

विकल्प

• \[\frac{\mu_0\pi r_1^2}{2r_2}\]

• \[\frac{\mu_0\pi r_1^2}{2r_1}\]

• \[\frac{\mu_0\pi\left(r_1+r_2\right)^2}{2r_1}\]

• \[\frac{\mu_0\pi\left(r_1-r_2\right)^2}{2r_2}\]

Answer 1

Two planar concentric rings of metal wire having radii r and r2 (r₁ > r2) are placed in air. The current I is flowing through the coil of larger radius. The mutual inductance between the coils is given by ( μο = permeability of free space) \[\frac{\mu_0\pi r_1^2}{2r_1}\].

Explanation:

The magnetic field at the centre of a loop is given by B = \[\frac{\mu_{0}\mathrm{NI}}{2\mathrm{R}}\]

\[\therefore\] Magnetic field produced by ring A, B_A = \[\frac{\mu_0\mathrm{I}}{2\mathrm{r}_1}\]

\[\therefore\] Magnetic flux produced in ring B due to B_A, 𝜙_B = B_A A_B

A_B = π\[{r}_{2}^{2}\]

\[\therefore\]\[\phi_\mathrm{B}=\frac{\mu_0\mathrm{I}}{2\mathrm{r}_1}\times\pi\mathrm{r}_2^2=\frac{\mu_0\pi\mathrm{r}_2^2}{2\mathrm{r}_1}\mathrm{I}\]

Mutual Inductance M = \[\frac {\phi}{I}\]

\[\therefore\] We can write, M = \[\frac{\phi_{\mathrm{B}}}{\mathrm{I}}=\frac{\mu_{0}\pi\mathrm{r}_{2}^{2}.\mathrm{I}}{2\mathrm{r}_{1}.\mathrm{I}}=\frac{\mu_{0}\pi\mathrm{r}_{2}^{2}}{2\mathrm{r}_{1}}\]