## Related questions with answers

Suppose the magnetic field of the preceding problem oscillates with time according to $B = B _ { 0 } \sin \omega t$. 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?

Solution

Verified#### Solution

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

$B = B_{o} \sin \omega t$

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

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

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

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

$B$ 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 $A = ld$ and equation (2) will be in the form

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

Where $l$ and $d$ are the sides of the rectangular loop. Now let us take the derivative of equation (1) for $B$ because it is the changeable parameter

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

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

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

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