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A charged particle of charge \(q\) and mass \(m\) is placed at a distance \(2 R\) from the centre of a vertical cylindrical region of radius \(R\) where magnetic field varies as \(\vec{B}=\left(4 t^2-2 t+6\right) \hat{k}\) where \(t\) is time. Then which of the following statement(s) is/are true?
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Verified Answer
The correct answers are:
Induced electric field lines form closed loops, Electric field varies linearly with \(r\) if \(r < R\), where \(r\) is the radial distance from the centerline of the cylinder
Hint : 
\(r < R\)
\(E \times 2 \pi r=\frac{d \phi}{d t}=\frac{d}{d t}\left(4 t^2-2 t+6\right) \times \pi r^2\)
\(\begin{aligned}
& E \times 2 \pi r=(8 t-2) \pi r^2 \\
& E=\frac{(8 t-2) r}{2} \\
& E=(4 t-1) r
\end{aligned}\)
\(E \propto r\)
For \(r > R\)
\(\begin{aligned}
& E \times 2 \pi \cdot(2 R)=\frac{d}{d t}\left[4 t^2-2 t+6\right] \times \pi[R]^2 \\
& E .4 \pi R=[8 t-2] \pi R^2 \\
& E=\frac{[8 t-2] R}{4} \quad \text { at } t=2, E=\frac{14}{4} R=\frac{7 R}{2} \\
& \text { acceleration }=\frac{E q}{m}=\frac{7 R q}{2 m}=\frac{7 q R}{2 m}
\end{aligned}\)
\(r < R\)
\(E \times 2 \pi r=\frac{d \phi}{d t}=\frac{d}{d t}\left(4 t^2-2 t+6\right) \times \pi r^2\)
\(\begin{aligned}
& E \times 2 \pi r=(8 t-2) \pi r^2 \\
& E=\frac{(8 t-2) r}{2} \\
& E=(4 t-1) r
\end{aligned}\)
\(E \propto r\)
For \(r > R\)
\(\begin{aligned}
& E \times 2 \pi \cdot(2 R)=\frac{d}{d t}\left[4 t^2-2 t+6\right] \times \pi[R]^2 \\
& E .4 \pi R=[8 t-2] \pi R^2 \\
& E=\frac{[8 t-2] R}{4} \quad \text { at } t=2, E=\frac{14}{4} R=\frac{7 R}{2} \\
& \text { acceleration }=\frac{E q}{m}=\frac{7 R q}{2 m}=\frac{7 q R}{2 m}
\end{aligned}\)
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