Question

Two helical coils of the same length but with different radii are arranged such that one is placed inside the other and both are co-aligned along the same central axis running along their length. They share a mutual inductance which can vary as a function of time by sliding the inner coil out of the outer coil. Calculate the EMF induced in the outer coil at the instant when

i. A current changes through the inner coil at a rate of 5 A s−1 and the mutual inductance is 0.4 mH.

ii. The current through the inner coil is 2 A and changing at a rate of 5 A s−1 and the mutual inductance is 0.4 mH but changing at a rate of 30 mH s−1.

iii. The current through the inner coil is fixed at 2 A and the mutual inductance is changing at a rate of 30 mH s−1.

Answer 1

given M=mutual inductance

and $i_2$ is current in smaller coil.

we know that,

flux through outer coil $(\phi_1)=Mi_2$.............1

where M is mutual inductance.

and induced EMF is given by $E=-\frac{d\phi}{dt}$ ..................2

putting the values and solving

a) using1 and 2 we get

$E=-M\frac{di}{dt}$

- 0.4mH $\times 5$ A/s

= - 2.0mH A /s

b)here using 1 and 2 we get

$E=-(M\frac{di}{dt}+i_2\frac{dM}{dt})$

= -{(0.4 $\times$ 5 )+(2 $\times$ 30)}

= - 62 mH A/S

c)using 1 qnd 2 we get

$E=-i_2\frac{dM}{dt}$

=-2×30

=-60 mH A/s

(here -ve sign depict direction opposite to original flow of emf in smaller coil )