A 1000 -turn coil of wire $1.0\mathrm{cm}$in diameter is in a magnetic field that increases from $0.10\mathrm{T}$to $0.30\mathrm{T}$in $10\mathrm{ms}$. The axis of the coil is parallel to the field. What is the emf of the coil?

Electromotive force is used by the electromagnetic flowmeter, which is based on Faraday's law. The electromotive force symbol is What does the Electromotive Force Formula entail? The electromotive force formula is as follows: What is the EMF unit? The unit of electromotive force is the volt.

According to Faraday's law, the induced emf is the change in magnetic flux inside the loop, and it is given by equation (30.14) in the form

$\epsilon =\left|\frac{d{\mathrm{\Phi }}_{\mathrm{m}}}{dt}\right|$

Where ${\Phi }_{\mathrm{m}}$is the flux through the loop which is the amount of magnetic field that flows through a loop of area A and it is given by

${\mathrm{\Phi }}_{\mathrm{m}}=NBA$

Let us use this expression of ${\Phi }_{\mathrm{m}}$into equation (1) to get $\epsilon$by

$\epsilon =NA\frac{d\left(BA\right)}{dt}=NA\frac{dB}{dt}$

The area of the loop is calculated by

$A=\pi {\left(\frac{d}{2}\right)}^2=\pi {\left(\frac{0.01\mathrm{m}}{2}\right)}^2=7.85\times {10}^{-5}{\mathrm{m}}^2$

The magnetic field changes from$0.1\mathrm{T}$to $0.3\mathrm{T}$in time $dt=10\mathrm{ms}$, so we get this change with time by

$\frac{dB}{dt}=\frac{0.3\mathrm{T}-0.1\mathrm{T}}{10\times {10}^{-3}\mathrm{s}}=20\mathrm{T}/\mathrm{s}$

Now, we plug the values for N, A and d B / d t into equation (2) to get $\epsilon$ by

$\epsilon =NA\frac{dB}{dt}$

$=\left(1000\right)\left(7.85\times {10}^{-5}{\mathrm{m}}^2\right)(20\mathrm{T}/\mathrm{s})$

The emf of the coil$\epsilon =1.6\mathrm{V}$