91Ó°ÊÓ

When changing current in one circuit changes the magnetic flux in other circuit, an emf is induced in the latter circuit. This emf is directly proportional to the time rate of current in former circuit. The constant so introduced removing the proportionality is called mutual inductance.

${\mathit{Î\mu }}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathit{M}\frac{\mathbf{d}{\mathbf{i}}_{\mathbf{1}}}{\mathbf{d}\mathbf{t}}$

Where is the induced emf in second circuit, is the time rate of current flowing in first circuit and is the mutual inductance.

The amount of magnetic flux linked with the two coils is directly proportional to the mutual inductance and current change, which follows the below relation

${\mathrm{Ï•}}_{B2}=\frac{M}{N_2}{i}_1$

Whereis magnetic flux linked with second coil having turns when current is flowing in first coil.

(a)Two coils are wound around the same cylindrical form. When the current in the first coil is decreasing at a rate of $\frac{d{i}_1}{dt}=-0.242A/s$, the induced emf in the second coil has magnitude ${\mathrm{Î\mu }}_2=1.65×{10}^{-3}V$. The mutual inductance is then calculated as

$$\begin{gathered} M=-\frac{{\mathrm{Î\mu }}_2}{\frac{d{i}_1}{dt}} \\ M=\frac{1.65×{10}^{-3}V}{0.242A/s} \\ M=6.82×{10}^{-3}H \end{gathered}$$

Therefore, the mutual inductance of the pair of coils is$6.82×{10}^{-3}H$.

(b) If the second coil has$N_2=25$turns and current in the first coil is $i_1=1.20A$ then magnetic flux so linked with second coil is given as

$$\begin{gathered} {\mathrm{ϕ}}_{B2}=\frac{M}{N_2}{i}_1 \\ {\mathrm{ϕ}}_{B2}=\frac{6.82×{10}^{-3}H×1.20A}{25} \\ {\mathrm{ϕ}}_{B2}=3.27×{10}^{-4}Wb \end{gathered}$$

Therefore, Magnetic flux through each turn of second coil is$3.27×{10}^{-4}Wb$

(c) If the current in the second coil increases at a rate of$\frac{d{i}_2}{dt}=0.360A/s$and mutual inductance of pair of$M=6.82×{10}^{-3}H$coils isthen magnitude of induced emf in first coil is,

$$\begin{gathered} \left|{\mathrm{Î\mu }}_1\right|=M\frac{d{i}_2}{dt} \\ {\mathrm{Î\mu }}_1=6.82×{10}^{-3}×0.360 \\ {\mathrm{Î\mu }}_1=2.46×{10}^{-3}V \end{gathered}$$

Therefore, the magnitude of induced emf in first coil when the current in the second coil increases at a rate of $\$0.360A/s\$$, is role="math" $2.46×{10}^{-3}V$.