Question

A small block of mass $m$, having charge $q$ is placed on frictionless inclined plane making an angle $\theta$ with the horizontal. There exists a uniform magnetic field $B$ parallel to the inclined plane but perpendicular to the length of spring. If $m$ is slightly pulled on the inclined in downward direction and released, the time period of oscillation will be (assume that the block does not leave contact with the plane)

• A. $2\pi \sqrt{\frac{m}{k}}$
• B. $2\pi \sqrt{\frac{2m}{k}}$
• C. $2\pi \sqrt{\frac{qB}{k}}$
• D. $2\pi \sqrt{\frac{qB}{2k}}$

Answer 1

The correct option is A $2\pi \sqrt{\frac{m}{k}}$

There will be no effect of magnetic force on time period because the magnetic force will be perpendicular to the inclined plane.
So, time period will be same as that due to motion on an inclined plane.
But $T=2\pi \sqrt{\frac{m}{k}}$