Mutual inductance of two coils can be increased by

Alternating current of peak value $(\frac{2}{\pi}) \ A$ flows through the primary coil of a transformer. The coefficient of mutual inductance between the primary and secondary coil is $1 \ H$. The peak e.m.f. induced in the secondary coil is (Frequency of a.c. $= 50 \ Hz$) (in $V$)

What is the mutual inductance of a two-loop system as shown with centre separation $l$ where $l >> a$?

$A$ circular loop of radius $0.3 \, cm$ lies parallel to a much bigger circular loop of radius $20 \, cm$. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is $15 \, cm$. If a current of $20 \, A$ flows through the smaller loop,then the flux linked with the bigger loop is:

$A$ solenoid of length $0.30 \, m$ has $2000$ turns. The area of its cross-section is $1.2 \times 10^{-3} \, m^2$. Around its central section,a coil of $300$ turns is wound. If an initial current of $2 \, A$ in the solenoid is reversed in $0.25 \, s$,then the $e.m.f.$ induced in the coil is:

$X$ and $Y$ are two circuits having a coefficient of mutual inductance $3 \text{ mH}$ and resistances $10 \text{ } \Omega$ and $4 \text{ } \Omega$ respectively. To have an induced current of $60 \times 10^{-4} \text{ A}$ in circuit $Y$,the amount of current to be changed in circuit $X$ in $0.02 \text{ s}$ is: (in $A$)