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# A 50-turn coil has a diameter of 15 cm. The coil is placed in a spatially uniform magnetic field of magnitude 0.50 T so that the face of the coil and the magnetic field are perpendicular. Find the magnitude of the emf induced in the coil if the magnetic field is reduced to zero uniformly in (a) 0.10 s, (b) 1.0 s, and (c) 60 s.

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#### Given

We are given the number of turns $N = 50$ and the diameter is $d$ =15 cm. The magnetic field through the coil is $B$ =0.50 T. The area is given by

\begin{align*} A= \pi \left(\dfrac{d}{2}\right)^{2}= \pi \left(\dfrac{15 \times 10^{-2}\,\text{m}}{2}\right)^{2} = 0.0176 \mathrm{~m^{2}} \end{align*}

#### Solution

(a) We want to determine the induced emf $\varepsilon$. When the coil moves in the magnetic field, an induced emf generated and it is given by equation 13.2 in the form

$$$\varepsilon = -N \dfrac{ d\Phi _{m}}{dt}$$$

Where $\Phi_{m}$ is the magnetic flux, which represents the number of magnetic field lines through the surface of the coil and it is given by equation 13.1 in the form

$\begin{equation*} \Phi_{m}=BA \tag{2} \end{equation*}$

The lines of the magnetic field are perpendicular to the surface, therefore equation 13.1 will be as shown in equation (2) where the field lines are parallel to the normal $\hat{n}$ of the coil.

Let us plug the expression of $\Phi_{m}$ into equation (1), where equation (1) will be in the form

\begin{align*} \varepsilon & =\left|\frac{-N d\Phi_{m}}{dt}\right| =\left|\frac{-N dBA}{dt}\right|= \left|\frac{-N BA}{t}\right| \tag{1*} \end{align*}

Now let us plug our values for $N$,$B$,$A$ and $t$ into equation (1*) to get $\varepsilon$

\begin{align*} \varepsilon & = \left|\frac{-N BA}{t}\right| = \left|\frac{-(50)(0.5 \mathrm{~T})(0.0176 \mathrm{~m^{2}})}{0.1 \,\text{s}}\right|= \boxed{4.4 \mathrm{~V}} \end{align*}

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