Respuesta :
Answer:
(A) Mutual inductance [tex]M=6.93\times 10^{-3}H[/tex]
(B) Flux will be [tex]0.275\times 10^{-3}weber[/tex]
(C) Induced emf will be [tex]2.529\times 10^{-3}volt[/tex]
Explanation:
We have given current in the first coil is decreasing as [tex]\frac{di}{dt}=0.245A/sec[/tex]
Emf induced [tex]e=1.70\times 10^{-3}volt[/tex]
(A) We know that emf induced is given by
[tex]e=M\frac{di}{dt}[/tex]
So [tex]1.70\times 10^{-3}=M\times 0.245[/tex]
[tex]M=6.93\times 10^{-3}H[/tex]
(B) Number of turns in second coil is N = 29
Current in first coil is 1.15 A
We know that [tex]N\Phi =Mi[/tex]
So [tex]\Phi =\frac{Mi}{N}=\frac{6.93\times 10^{-3}\times 1.15}{29}=0.275\times 10^{-3}weber[/tex]
(c) Rate of change of current in the second coil [tex]\frac{di}{dt}=0.365A/sec[/tex]
Induced emf is equal to [tex]e=M\frac{di}{dt}=6.93\times 10^{-3}\times 0.365=2.529\times 10^{-3}volt[/tex]
(a) The mutual inductance of the pair of coils is 6.9 x 10⁻³ H.
(b) The flux through each turn is 2.74 x 10⁻⁴ Wb.
(c) The magnitude of induced emf in the second coil 2.52 x 10⁻³ V.
The given parameters;
- change in current, di/dt = 0.245 A/s
- magnitude of induced emf, = 1 .7 x 10⁻³ V
The mutual inductance of the pair of coils is calculated as follows;
[tex]emf = M\frac{di}{dt} \\\\1.7\times 10^{-3} = M(0.245)\\\\M = \frac{1.7\times 10^{-3}}{0.245} \\\\M = 6.9 \times 10^{-3} \ H[/tex]
The flux through each turn when the current in the first coil equals 1.15 A and total equals 29, is calculated as;
[tex]N\Phi = Mi\\\\\Phi = \frac{Mi}{N} \\\\\Phi = \frac{6.9 \times 10^{-3} \times 1.15}{29} \\\\\Phi = 2.74 \times 10^{-4} \ Wb[/tex]
The magnitude of induced emf in the second coil when the current increases by 0.365 A/s.
[tex]emf = M\frac{di}{dt} \\\\emf = 6.9 \times 10^{-3} \times 0.365\\\\emf = 2.52 \times 10^{-3} \ V[/tex]
Learn more here:https://brainly.com/question/13154392
