Question 45: Electromagnetic Induction

Question

Consider I
1
and I
2
are the currents flowing simultaneously in two nearby coils 1 & 2, respectively. If L
1
= self inductance of coil 1, M
12
= mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be :

Accepted Answer

$$\begin{aligned}
& \phi_1=\mathrm{L}_1 \mathrm{I}_1+\mathrm{M}_{12} \mathrm{I}_2 \
& \varepsilon_1=-\frac{\mathrm{d} \phi_1}{\mathrm{dt}}=-\mathrm{L}_1 \frac{\mathrm{dI}_1}{\mathrm{dt}}-\mathrm{M}_{12} \frac{\mathrm{dI}}{\mathrm{dt}}
\end{aligned}$$

Answer

e
1
= -L
1
$\frac{dI_2}{dt}$ - M
12
$\frac{dI_1}{dt}$

Answer

e
1
= -L
1
$\frac{dI_1}{dt}$ + M
12
$\frac{dI_2}{dt}$

Answer

e
1
= -L
1
$\frac{dI_1}{dt}$ - M
12
$\frac{dI_1}{dt}$

Answer

e
1
= -L
1
$\frac{dI_1}{dt}$ - M
12
$\frac{dI_2}{dt}$

Electromagnetic Induction

Consider I 1 and I 2 are the currents flowing simultaneously in two nearby coils 1 & 2, respectively. If L 1 = self inductance of coil 1, M 12 = mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be :

Choose the correct answer: