Question

In an electromagnetics lab, you are studying two coils which have a mutual inductance of M=300 mH. Suppose that the current in the 1st coil increased linearly from 2.8 A to 10 A in a span of 300 ms and that the 2nd coil has a resistance of R=0.4 Ohm, what is the magnitude of the induced current I in the 2nd coil? (Note: write only the two digits in the space provided for value of current in amperes).

Answer 1

Answer:

Explanation:

Given that,

The mutual inductance of the two coils is

M = 300mH = 300 × 10^-3 H

M = 0.3 H

Current increase in the coil from 2.8A to 10A

∆I = I_2 - I_1 = 10 - 2.8

∆I = 7.2 A

Within the time 300ms

t = 300ms = 300 × 10^-3

t = 0.3s

Second Coil resistance

R_2 = 0.4 ohms

We want to find the current in the second coil,

The same induced EMF is in both coils, so let find the EMF,

From faradays law

ε = Mdi/dt

ε = M•∆I / ∆t

ε = 0.3 × 7.2 / 0.3

ε = 7.2 Volts

Now, this is the voltage across both coils,

Applying ohms law to the second coil, V=IR

ε = I_2•R_2

0.72 = I_2 • 0.4

I_2 = 0.72 / 0.4

I_2 = 1.8 Amps

The current in the second coil is 1.8A