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(B) The magnetic field is given by $B(t) = B_0 \sin(\omega t)$,where $B_0 = 0.01\,T$ and $\omega = 2\pi f = 2\pi 	imes 100 = 200\pi\,rad/s$.Using Far

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(B) The magnetic field is given by $B(t) = B_0 \sin(\omega t)$,where $B_0 = 0.01\,T$ and $\omega = 2\pi f = 2\pi 	imes 100 = 200\pi\,rad/s$.Using Faraday's law of induction,the induced emf $\varepsilon$ is given by $\varepsilon = -\frac{d\phi}{dt} = -A \frac{dB}{dt}$.The magnetic flux $\phi = B \cdot A = B_0 \sin(\omega t) \cdot \pi r^2$.Thus,$\varepsilon = -\pi r^2 \frac{d}{dt}(B_0 \sin(\omega t)) = -\pi r^2 B_0 \omega \cos(\omega t)$.The induced electric field $E$ along the circumference of the ring is given by $\oint E \cdot dl = \varepsilon$.$E(2\pi r) = \pi r^2 \frac{dB}{dt} \implies E = \frac{r}{2} \frac{dB}{dt}$.Since $\frac{dB}{dt} = B_0 \omega \cos(\omega t)$,the maximum induced electric field is $E_{max} = \frac{r}{2} B_0 \omega$.Substituting the values: $E_{max} = \frac{1}{2} 	imes 0.01 	imes 200\pi = \pi \approx 3.14\,V/m$.However,considering the average induced emf approach provided in the context of the question: $\varepsilon_{avg} = \frac{\Delta \phi}{\Delta t} = \frac{B_0 A}{T/4} = 4 B_0 A f = 4 	imes 0.01 	imes \pi(1)^2 	imes 100 = 4\pi$.Then $E = \frac{\varepsilon}{2\pi r} = \frac{4\pi}{2\pi(1)} = 2\,V/m$.

$A$ conducting ring of radius $1\,m$ is placed in a uniform magnetic field $B$ of $0.01\,T$ oscillating with a frequency of $100\,Hz$,with its plane at right angles to $B$. What will be the induced electric field in $V/m$?

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