Two coils have a mutual inductance of $0.005\, ext{H}$. The current in the first coil changes according to $i=I_0\sin\omega t$, where $I_0=10\, ext{A}$ and $\omega=100\pi\, ext{rad s}^{-1}$. The maximum value of e.m.f. in the second coil is:

Question

Two coils have a mutual inductance of $0.005\,	ext{H}$. The current in the first coil changes according to $i=I_0\sin\omega t$, where $I_0=10\,	ext{A}$ and $\omega=100\pi\,	ext{rad s}^{-1}$. The maximum value of e.m.f. in the second coil is:

Accepted Answer

$5\pi$. The induced emf in the second coil is $e=M\,\dfrac{di}{dt}$. For $i=I_0\sin\omega t$, the maximum rate of change is $\omega I_0$, so $e_{\max}=M\omega I_0=0.005	imes100\pi	imes10=5\pi\,	ext{V}$.

Two coils have a mutual inductance of $0.005\,\text{H}$. The current in the first coil changes according to $i=I_0\sin\omega t$, where $I_0=10\,\text{A}$ and $\omega=100\pi\,\text{rad s}^{-1}$. The maximum value of e.m.f. in the second coil is:

Solution

Related NEET Physics questions