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Question

Suppose the magnetic field of the preceding problem oscillates with time according to B=B0sinωtB = B _ { 0 } \sin \omega t. What then is the emf induced in the loop when its trailing side is a distance d from the right edge of the magnetic field region?

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Solution

We are given that the magnetic field varies with time with the next equation

B=BosinωtB = B_{o} \sin \omega t

According to Faraday's law, the induced emf is given by

ε=NdΦmdt\begin{equation*} \varepsilon= - N \frac{d\Phi_{m}}{dt} \tag{1} \end{equation*}

Where N=1N = 1 and Φm\Phi_{m} is the magnetic flux in the loop and it is given by equation 13.1 in the form

Φm=Bn^dA\begin{equation*} \Phi_{m}=\int \vec{B} \cdot \hat{n} dA \tag{2} \end{equation*}

BB is perpendicular to the plane of the loop, so the dot product will convert to a multiplication of magnetic field and area where the area, in this case, will be A=ldA = ld and equation (2) will be in the form

Φm=BA=Bld\begin{equation*} \Phi_{m} = BA = Bld \tag{3} \end{equation*}

Where ll and dd are the sides of the rectangular loop. Now let us take the derivative of equation (1) for BB because it is the changeable parameter

ε=dΦmdt=dBlddt=lddBdt\begin{equation*} \varepsilon = - \frac{d\Phi_{m}}{dt} = \frac{dBld}{dt} = ld \frac{dB}{dt} \tag{1*} \end{equation*}

Now let us use the expression of BB and take the derivative for time to get the expression of ε\varepsilon

ε=ldd(Bosinωt)dt=ldBoωcosωt\begin{align*} \varepsilon = - ld \frac{d (B_{o} \sin \omega t) }{dt} = \boxed{- ldB_{o}\omega \cos \omega t} \end{align*}

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