JEE MAIN - Physics (2024 - 31st January Morning Shift - No. 13)
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:
35
70
7
140
Explanation
$$\begin{aligned}
\varepsilon & =\mathrm{N}\left(\frac{\Delta \phi}{\mathrm{t}}\right) \\
\Delta \phi & =(\Delta \mathrm{B}) \mathrm{A} \\
\mathrm{B}_{\mathrm{i}} & =5000 \mathrm{~T}, \\
\mathrm{~B}_{\mathrm{f}} & =3000 \mathrm{~T} \\
\mathrm{~d} & =0.02 \mathrm{~m} \\
\mathrm{r} & =0.01 \mathrm{~m} \\
\Delta \phi & =(\Delta \mathrm{B}) \mathrm{A} \\
& =(2000) \pi(0.01)^2=0.2 \pi \\
\varepsilon & =\mathrm{N}\left(\frac{\Delta \phi}{\mathrm{t}}\right) \Rightarrow 22=\mathrm{N}\left(\frac{0.2 \pi}{2}\right) \\
\mathrm{N} & =70
\end{aligned}$$
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