Question

A charged particle of charge q and mass m is placed at a distance 2R from the center of a vertical cylindrical region of radius R where the magnetic field varies as \(\vec{B}=(4t^2-2t+6)k\), where t is time. Then which of the following statement(s) is/are true?

• A. Induced electric field lines from closed loops
• B. The Electric field varies linearly with r if r< R, where r is the radial distance from the centerline of the cylinder.
• C. The charged particle will move in a clockwise direction when viewed from the top.
• D. Acceleration of the charged particle is \(\frac{7q}{2m}\) when t=2 sec.

Answer 1

The Correct Option is A, B, C

Step 1: Magnetic Field and Induced Electric Field

The magnetic field inside the cylindrical region is given as: $$ \mathbf{B} = (4t^2 - 2t + 6)\hat{k} $$ where \( t \) is time. Since the magnetic field varies with time, an induced electric field is generated according to Faraday's Law of Electromagnetic Induction: $$ \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} $$ This implies that the induced electric field forms closed loops because the curl of the electric field is non-zero.

Thus, option (A) is correct.

Step 2: Electric Field Inside the Cylinder (\( r < R \))

For \( r < R \), the induced electric field can be determined using Faraday's Law. The rate of change of the magnetic flux through a loop of radius \( r \) is: $$ \frac{d\Phi_B}{dt} = \pi r^2 \frac{\partial B}{\partial t} $$ Substituting \( B = 4t^2 - 2t + 6 \): $$ \frac{\partial B}{\partial t} = 8t - 2 $$ Therefore: $$ \frac{d\Phi_B}{dt} = \pi r^2 (8t - 2) $$ By Faraday's Law, the induced electric field satisfies: $$ \oint \mathbf{E} \cdot d\mathbf{l} = -\frac{d\Phi_B}{dt} $$ For a circular loop of radius \( r \), the electric field magnitude is: $$ E \cdot (2\pi r) = -\pi r^2 (8t - 2) $$ Simplifying: $$ E = -\frac{r}{2} (8t - 2) $$ Thus, the electric field \( E \) varies linearly with \( r \) for \( r < R \).

This confirms that option (B) is correct.

Step 3: Direction of Motion of the Charged Particle

At \( t = 2 \, \text{sec} \), the magnetic field is: $$ B = 4(2)^2 - 2(2) + 6 = 16 - 4 + 6 = 18 \, \text{T} $$ The rate of change of the magnetic field is: $$ \frac{\partial B}{\partial t} = 8t - 2 = 8(2) - 2 = 16 - 2 = 14 \, \text{T/s} $$ The induced electric field creates a force on the charged particle. Using the right-hand rule, the direction of the induced electric field is such that it opposes the change in magnetic flux. When viewed from the top, the induced electric field causes the charged particle to move in a clockwise direction.

This confirms that option (C) is correct.

Step 4: Acceleration of the Charged Particle

The force on the charged particle due to the induced electric field is: $$ F = qE $$ At \( r = 2R \), the electric field is: $$ E = -\frac{r}{2} (8t - 2) = -\frac{2R}{2} (8t - 2) = -R(8t - 2) $$ Substituting \( t = 2 \, \text{sec} \): $$ E = -R(8(2) - 2) = -R(16 - 2) = -14R $$ The force is: $$ F = q(-14R) = -14qR $$ The acceleration is: $$ a = \frac{F}{m} = \frac{-14qR}{m} $$ This does not match the given value of \( \frac{7q}{2m} \). Thus, option (D) is incorrect.

Final Answer:

The correct option(s) is/are: $$ \boxed{\text{(A), (B), and (C)}} $$

Answer 2

Correct statements: (A), (B), and (C) 

Given: 
- A charged particle with charge $q$ and mass $m$ is at a distance $2R$ from the center of a vertical cylindrical region of radius $R$.
- Magnetic field: $B = (4t^2 - 2t + 6) \hat{k}$ 

Using Faraday's Law:
\[ \oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt} \] \[ \Phi_B = B \cdot \pi r^2 = (4t^2 - 2t + 6) \cdot \pi r^2 \] \[ \mathcal{E} = -\frac{d\Phi_B}{dt} = -\pi r^2 (8t - 2) \] \[ E \cdot 2\pi r = -\pi r^2 (8t - 2) \] \[ E = -\frac{r}{2}(8t - 2) \] 

Now analyzing the options:
(A) The induced electric field lines form closed loops. 
✅ True, as per Faraday's law, the induced $\vec{E}$ forms circular closed loops. 

(B) The electric field varies linearly with $r$ if $r < R$. 
✅ True, from $E = -\frac{r}{2}(8t - 2)$, it is directly proportional to $r$. 

(C) The charged particle moves in a clockwise direction when viewed from the top. 
✅ True, magnetic field is increasing (since $dB/dt = 8t - 2$ is positive for $t > 0.25$), ⇒ induced field is clockwise to oppose this (Lenz’s Law) ⇒ positive charge moves clockwise. 

Conclusion:
✅ Correct options are (A), (B), and (C).