Question

A smooth cylinder is fixed on a smooth surface with its axis vertical and radius 1 cm. If a downward magnetic field of magnetic field intensity $1 \times 10^5$ T exists in a region and charge (q) = $100\mu$C of mass 100 gm is projected with velocity 4 m/s in anticlockwise direction then:

• A. Normal exerted by the walls of the cylinder is 120 N
• B. Normal exerted by walls of cylinder is zero
• C. Particle will fly radially outward when it reaches position A on the cylinder.
• D. Radius of curvature at position A will be 4 cm

Answer 1

Answer: (A), (D)

A, D

Answer 2

The problem describes a charged particle moving inside a smooth cylinder in a uniform magnetic field. We need to analyze the forces acting on the particle and determine its motion.

1. Calculate Magnetic Force ($F_B$): $F_B = qvB = (10^{-4} \text{ C})(4 \text{ m/s})(1 \times 10^5 \text{ T}) = 40 \text{ N}$.
2. Determine Direction of $F_B$: Using the right-hand rule for $\vec{F}_B = q(\vec{v} \times \vec{B})$, with $\vec{v}$ anticlockwise and $\vec{B}$ downward (into page), $\vec{F}_B$ is radially inward.
3. Calculate Required Centripetal Force ($F_c$): $F_c = \frac{mv^2}{R} = \frac{(0.1 \text{ kg})(4 \text{ m/s})^2}{0.01 \text{ m}} = 160 \text{ N}$.
4. Calculate Normal Force ($N$): Both $N$ and $F_B$ are radially inward. So, $N + F_B = F_c \implies N = F_c - F_B = 160 \text{ N} - 40 \text{ N} = 120 \text{ N}$. (Option A is correct).
5. Calculate Radius of Curvature at Position A ($r'$): At the hole, $N=0$. Only $F_B$ provides the centripetal force. $F_B = \frac{mv^2}{r'} \implies qvB = \frac{mv^2}{r'} \implies r' = \frac{mv}{qB}$.
6. Substitute values for $r'$: $r' = \frac{(0.1 \text{ kg})(4 \text{ m/s})}{(10^{-4} \text{ C})(1 \times 10^5 \text{ T})} = \frac{0.4}{10} = 0.04 \text{ m} = 4 \text{ cm}$. (Option D is correct).
7. Check Option C: Since $F_B$ is inward, the particle will curve inward, not fly radially outward. (Option C is incorrect).