A coil is places perpendicular to a magnetic field of $5000 \mathrm{~T}$. When the field is changed to $3000 \mathrm{~T}$ in $2 \mathrm{~s}$, an induced emf of $22 \mathrm{~V}$ is produced in the coil. If the diameter of the coil is $0.02 \mathrm{~m}$, then the number of turns in the coil is:
Solution
<p>$$\begin{aligned}
\varepsilon & =\mathrm{N}\left(\frac{\Delta \phi}{\mathrm{t}}\right) \\
\Delta \phi & =(\Delta \mathrm{B}) \mathrm{A} \\
\mathrm{B}_{\mathrm{i}} & =5000 \mathrm{~T}, \\
\mathrm{~B}_{\mathrm{f}} & =3000 \mathrm{~T} \\
\mathrm{~d} & =0.02 \mathrm{~m} \\
\mathrm{r} & =0.01 \mathrm{~m} \\
\Delta \phi & =(\Delta \mathrm{B}) \mathrm{A} \\
& =(2000) \pi(0.01)^2=0.2 \pi \\
\varepsilon & =\mathrm{N}\left(\frac{\Delta \phi}{\mathrm{t}}\right) \Rightarrow 22=\mathrm{N}\left(\frac{0.2 \pi}{2}\right) \\
\mathrm{N} & =70
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Electromagnetic Induction · Topic: Faraday's Laws
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