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 , \(\mathrm{M}_{12}=\) mutual inductance of coil 1 with respect to coil 2 , then the value of induced emf in coil 1 will be

(1) \(\varepsilon_{1}=-\mathrm{L}_{1} \frac{\mathrm{dI_1}}{\mathrm{dt}}+\mathrm{M}_{12} \frac{\mathrm{dI} I_{2}}{\mathrm{dt}}\)

(2) \(\varepsilon_{1}=-\mathrm{L}_{1} \frac{\mathrm{dI}_{1}}{\mathrm{dt}}-\mathrm{M}_{12} \frac{\mathrm{dI}_{1}}{\mathrm{dt}}\)

(3) \(\varepsilon_{1}=-\mathrm{L}_{1} \frac{\mathrm{dI}_{1}}{\mathrm{dt}}-\mathrm{M}_{12} \frac{\mathrm{dI} I_{2}}{\mathrm{dt}}\)

(4) \(\varepsilon_{1}=-\mathrm{L}_{1} \frac{\mathrm{dI}_{2}}{\mathrm{dt}}-\mathrm{M}_{12} \frac{\mathrm{dI}_{1}}{\mathrm{dt}}\)

Answer 1

Correct option is (3) \(\varepsilon_{1}=-\mathrm{L}_{1} \frac{\mathrm{dI}_{1}}{\mathrm{dt}}-\mathrm{M}_{12} \frac{\mathrm{dI} I_{2}}{\mathrm{dt}}\)   

\(\phi_{1}=L_{1} I_{1}+M_{12} I_{2}\)

\(\varepsilon_{1}=-\frac{\mathrm{d} \phi_{1}}{\mathrm{dt}}=-\mathrm{L}_{1} \frac{\mathrm{dI}_{1}}{\mathrm{dt}}-\mathrm{M}_{12} \frac{\mathrm{dI}_{2}}{\mathrm{dt}}\)