A charged particle of charge q and mass m is placed at a distance 2R from the centre of a vertical cylindrical region of radius R where magnetic field varies as \(\vec{B}=(4t^{2}-2t+6)\hat{k}\), where t is time. Then which of the following statement(s) is/are true ?
A time-varying magnetic field produces an induced electric field even where the magnetic field itself is zero. By Faraday's law in integral form, \(\oint\vec{E}\cdot d\vec{l}=-\dfrac{d\Phi_{B}}{dt}\); by symmetry the induced field is tangential and has the same magnitude on any circle coaxial with the cylinder.
Step 1 — Inside the region (r < R): \(E(2\pi r)=\pi r^{2}\dfrac{dB}{dt}\;\Rightarrow\;E=\dfrac{r}{2}\dfrac{dB}{dt}\), so the field grows linearly with r.
Step 2 — Outside the region (r > R) the whole flux πR2B is enclosed: \(E(2\pi r)=\pi R^{2}\dfrac{dB}{dt}\;\Rightarrow\;E=\dfrac{R^{2}}{2r}\dfrac{dB}{dt}\).
Step 3 — At t = 2 s, \(\dfrac{dB}{dt}=8t-2=14\ \text{T s}^{-1}\). At r = 2R this gives \(E=\dfrac{R^{2}}{2(2R)}(14)=\dfrac{7R}{2},\quad a=\dfrac{qE}{m}=\dfrac{7qR}{2m}\), which contains the factor R and is not 7q/2m.
Step 4 — For t > 1/4 s, dB/dt = 8t − 2 is positive, so the field along +k̂ (upwards) is increasing, so the induced electric field circulates to oppose the increase: clockwise when viewed from the top. A positive charge is pushed along this direction.
The induced electric field is non-conservative and its field lines close on themselves, so the closed-loop statement holds. The linear variation inside the cylinder and the clockwise sense viewed from the top also hold. The acceleration statement fails because the correct value 7qR/2m depends on R and does not reduce to 7q/2m.
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. and The charged particle will move in clockwise direction when viewed from top. options are correct.
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