A conducting disc of mass ‘m’ and volume ‘V’ is suspended with the help of a light spring of force constant ‘k’ from a fixed support. Thickness of the disc is much smaller than its radius. A uniform magnetic field of induction ‘B’ parallel to the plane of the disc is established. The disc is slightly pulled down from equilibrium position and then released. If the time period of small oscillations of the disc is 2παmk−−−√. Find the value of α. (Take ε0VB2=2m)
Answers
Answer:
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Given:
Mass of the conducting disc: m
Volume of the conducting disc: V
Force constant of the light spring: k
Thickness of the disc << radius
Uniform magnetic field induction parallel to the plane of the disc: B
Time period of small oscillations of the disc: 2παmk^(-1/2)
ε0VB^2 = 2m
To find: the value of α in the given scenario
Solution:
For a simple harmonic motion of a spring-block system, the time period is given by T = 2π(m/k)^0.5.
In the given scenario, the disc is performing a simple harmonic motion and the force acting on it is the magnetic force due to the magnetic field. Therefore, the restoring force is proportional to the displacement of the disc from its equilibrium position.
The magnetic force acting on the disc is given by the formula F = B^2 * V * ε0 * h^2 / (2 * m), where h is the thickness of the disc.
The restoring force can be calculated as F = k * x, where x is the displacement from the equilibrium position.
Equating the two forces, we get:
k * x = B^2 * V * ε0 * h^2 * x / (2 * m)
Simplifying and solving for k, we get:
k = B^2 * V * ε0 * h^2 / (2 * m)
Using the given formula for time period, we get:
2παmk^(-1/2) = 2π(m/k)^0.5
Simplifying and solving for α, we get:
α = hB^2/(4πε0mk)
Therefore, the value of α in the given scenario is hB^2/(4πε0mk).
To learn more about Mass from the given link.
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