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
24hours is the correct answer
Answer:
The induced current is 26.7 mA
Explanation:
Given;
area of the loop, A = 0.078 m²
initial magnetic field, B₁ = 3.8 T
change in the magnetic field strength, dB/dt = 0.24 T/s
The induced emf is calculated as;
The resistance of the loop = 0.7 Ω
The induced current is calculated as;
