A small block of mass m,having charge q,is pl | vedclass

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(A) The magnetic force acting on the moving charge $q$ is given by $\vec{F}_m = q(\vec{v} 	imes \vec{B})$.Given that the magnetic field $\vec{B}$ is

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$2\pi \sqrt{\frac{m}{k}}$

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(A) The magnetic force acting on the moving charge $q$ is given by $\vec{F}_m = q(\vec{v} 	imes \vec{B})$.Given that the magnetic field $\vec{B}$ is parallel to the inclined plane and the velocity $\vec{v}$ of the block is also along the inclined plane,the magnetic force will be perpendicular to the inclined plane.Since the magnetic force acts perpendicular to the plane of motion,it does not affect the restoring force of the spring,which acts along the inclined plane.The equation of motion for the block along the inclined plane is $m \frac{d^2x}{dt^2} = -kx$,where $x$ is the displacement from the equilibrium position.This is the standard equation for simple harmonic motion with angular frequency $\omega = \sqrt{\frac{k}{m}}$.The time period of oscillation is $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{k}}$.Thus,the magnetic field has no effect on the time period of oscillation.

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