Question
Explain mutual induction using Faraday's two coil experiment. Obtain expression
✓
Answer
$\varepsilon_1=- M \frac{d l_2}{d t}$.
→In the figure, arrangement of two coil experiment is shown. Here, coils $C _1$ and $C _2$ are placed coaxially.
→When electric current $I _2$ passing through coil $C _2$ changes, magnetic flux linked with coil $C _1$ changes and hence induced emf $\varepsilon_1$ arises.
→When electric current $I _2$ flows through $C _2$, total magnetic flux linked with coil $C _1$ having $N _1$ turns is $N _1 \phi_1 \propto I _2$
$\therefore N _1 \phi_1= M I _2$
Where, M is mutual inductance
→For current changing with time,
$\frac{d}{d t}\left(N_1 \phi_1\right)= M \frac{d}{d t}\left( I _2\right)$
→According to Faraday's law, induced emf in $C _1$
$\varepsilon_1=-\frac{d}{d t}\left(N_1 \phi_1\right)$
→From equation (2) and (3),
$\varepsilon_1=- M \frac{d I _2}{d t}$
→Equation (4) is the required result.
→Here, value of induced emf depends on time rate of change of electric current and mutual inductance.
→In the figure, arrangement of two coil experiment is shown. Here, coils $C _1$ and $C _2$ are placed coaxially.
→When electric current $I _2$ passing through coil $C _2$ changes, magnetic flux linked with coil $C _1$ changes and hence induced emf $\varepsilon_1$ arises.
→When electric current $I _2$ flows through $C _2$, total magnetic flux linked with coil $C _1$ having $N _1$ turns is $N _1 \phi_1 \propto I _2$
$\therefore N _1 \phi_1= M I _2$
Where, M is mutual inductance
→For current changing with time,
$\frac{d}{d t}\left(N_1 \phi_1\right)= M \frac{d}{d t}\left( I _2\right)$
→According to Faraday's law, induced emf in $C _1$
$\varepsilon_1=-\frac{d}{d t}\left(N_1 \phi_1\right)$
→From equation (2) and (3),
$\varepsilon_1=- M \frac{d I _2}{d t}$
→Equation (4) is the required result.
→Here, value of induced emf depends on time rate of change of electric current and mutual inductance.
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