Question

A rectangular loop of area A is placed in a region where the magnetic field is perpendicular to the plane of the loop. The magnitude of the field is allowed to vary in time according to B = Bmax e-t/τ , where Bmax and τ are constants. The field has the constant value Bmax for t < 0. Find the emf induced in the loop as a function of time. (Use the following as necessary: A, Bmax, t, and τ.) g

Answer 1

Answer:

Induced emf, $\epsilon=-A\dfrac{B_{max}e^{-t/\tau}}{\tau}$

Explanation:

The varying magnetic field with time t is given by according to equation as :

$B=B_{max}e^{-t/\tau}$

Where

$B_{max}\ and\ t$ are constant

Let $\epsilon$ 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:

$\epsilon=-\dfrac{d\phi}{dt}$

$\epsilon=-\dfrac{d(BA)}{dt}$

$\epsilon=-A\dfrac{d(B)}{dt}$

$\epsilon=-A\dfrac{d(B_{max}e^{-t/\tau})}{dt}$

$\epsilon=A\dfrac{B_{max}e^{-t/\tau}}{\tau}$

So, the induced emf in the loop as a function of time is $A\dfrac{B_{max}e^{-t/\tau}}{\tau}$. Hence, this is the required solution.

Answer 2

The emf induced in the loop as a function of time is determined as $emf = \frac{AB_{max}}{\tau} e^{-t/\tau}$.

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

$emf = \frac{d(BA)}{dt} \\\\emf = A \frac{dB}{dt} \\\\emf = A \times B_{max}(e^{-t/\tau})(1/\tau)\\\\emf = \frac{A B_{max}(e^{-t/\tau})}{\tau}\\\\emf = \frac{AB_{max}}{\tau} e^{-t/\tau}$

Thus, the emf induced in the loop as a function of time is determined as $emf = \frac{AB_{max}}{\tau} e^{-t/\tau}$.

Learn more about induced emf here: https://brainly.com/question/13744192