Two coils A and B having turns 300 and 600 re | vedclass

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(B) The mutual inductance $M$ of a system of two coils is defined by the relation between the current in the primary coil $(i_1)$ and the flux linkage

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$3 	imes 10^{-5} \ H$

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(B) The mutual inductance $M$ of a system of two coils is defined by the relation between the current in the primary coil $(i_1)$ and the flux linkage of the secondary coil $(N_2 \phi_2)$:$N_2 \phi_2 = M i_1$Given:Number of turns in coil $B$ $(N_2)$ = $600$Flux linked with coil $B$ $(\phi_2)$ = $9.0 	imes 10^{-5} \ Wb$Current in coil $A$ $(i_1)$ = $3.0 \ A$Substituting the values into the formula:$600 	imes (9.0 	imes 10^{-5}) = M 	imes 3.0$$5400 	imes 10^{-5} = 3.0 	imes M$$5.4 	imes 10^{-2} = 3.0 	imes M$$M = \frac{5.4 	imes 10^{-2}}{3.0} = 1.8 	imes 10^{-2} \ H$Wait, checking the flux linkage definition: The flux linkage with coil $B$ is already given as $N_2 \phi_2 = 9.0 	imes 10^{-5} \ Wb$ (as the total flux linked with the coil is the product of turns and flux per turn).Using $N_2 \phi_2 = M i_1$:$9.0 	imes 10^{-5} = M 	imes 3.0$$M = 3.0 	imes 10^{-5} \ H$.

Two coils $A$ and $B$ having turns $300$ and $600$ respectively are placed near each other. On passing a current of $3.0 \ A$ in coil $A$, the magnetic flux linked with coil $A$ is $1.2 \times 10^{-4} \ Wb$ and with coil $B$ it is $9.0 \times 10^{-5} \ Wb$. The mutual inductance of the system is:

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