Question

Two coils of wire are placed close together. Initially, a current of 1.04 A exists in one of the coils, but there is no current in the other. The current is then switched off in a time of 2.23 x 10-2 s. During this time, the average emf induced in the other coil is 2.66 V. What is the mutual inductance of the two-coil system?

Answer 1

M=0.057H

Step-by-step explanation:

we have that both emf generated in two wires is given by:

`\epsilon_1=-M(dI_2)/(dt)\\\\\epsilon_2=-M(dI_1)/(dt)`

where M is the mutual inductance. We can use the second expression for emf2:

`2.66V=-M(\Delta I_1)/(\Delta t)=-M(0A-1.04A)/(2.23*10^(-2)s-0s)=M46.63(A)/(s)\\\\M=(2.66)/(46.63)H=0.057H`

hope this helps!!

Answer 2

Step-by-step explanation:

Given that,

Given two coils

Current in first coil

i1 = 1.04A

No current in the second coil

i2 = 0A

Current in switch turn off within

t = 2.23 × 10^-2 secs

Induced EMF in the coil

ε = 2.66V

Mutual inductance M?

The EMF in the coil can be calculated using

ε = —M•di/dt

Where

M is mutual inductance in Henry

ε is induced EMF in coils in Voltage

di/dt is rate of change of current in Amps/Sec

ε = —M• ∆I/∆t

Make M subject of formula

M = —ε∆t/∆I

M = —ε (t2—t1) / (I2—I1)

M =—2.66 × (2.23 × 10^-2 — 0) / (0-1.04)

M = —2.66 ×2.23 × 10^-2 / —1.04

M = 0.05704 H

M = 57.04 mH

The mutual inductance between the coil is 57.04mH