Respuesta :
Answer:
Induced emf, [tex]\epsilon=-A\dfrac{B_{max}e^{-t/\tau}}{\tau}[/tex]
Explanation:
The varying magnetic field with time t is given by according to equation as :
[tex]B=B_{max}e^{-t/\tau}[/tex]
Where
[tex]B_{max}\ and\ t[/tex] are constant
Let [tex]\epsilon[/tex] is the emf induced in the loop as a function of time. We know that the rate of change of magnetic flux is equal to the induced emf as:
[tex]\epsilon=-\dfrac{d\phi}{dt}[/tex]
[tex]\epsilon=-\dfrac{d(BA)}{dt}[/tex]
[tex]\epsilon=-A\dfrac{d(B)}{dt}[/tex]
[tex]\epsilon=-A\dfrac{d(B_{max}e^{-t/\tau})}{dt}[/tex]
[tex]\epsilon=A\dfrac{B_{max}e^{-t/\tau}}{\tau}[/tex]
So, the induced emf in the loop as a function of time is [tex]A\dfrac{B_{max}e^{-t/\tau}}{\tau}[/tex]. Hence, this is the required solution.
The emf induced in the loop as a function of time is determined as [tex]emf = \frac{AB_{max}}{\tau} e^{-t/\tau}[/tex].
Induced emf
The emf induced in the rectangular loop is determined by applying Faraday's law of electromagnetic induction.
emf = dФ/dt
where;
- Ф is magnetic flux
Ф = BA
A is the area of the rectangular loop
emf = d(BA)/dt
[tex]emf = \frac{d(BA)}{dt} \\\\emf = A \frac{dB}{dt} \\\\emf = A \times B_{max}(e^{-t/\tau})(1/\tau)\\\\emf = \frac{A B_{max}(e^{-t/\tau})}{\tau}\\\\emf = \frac{AB_{max}}{\tau} e^{-t/\tau}[/tex]
Thus, the emf induced in the loop as a function of time is determined as [tex]emf = \frac{AB_{max}}{\tau} e^{-t/\tau}[/tex].
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