Answer to Question #193357 in Electric Circuits for benjamin johnson

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 )