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

The spring elongates by $x_o$ unit from its natural length due to the weight of the block so that the block is in equilibrium about axis O.

$\therefore mgsin\theta =k{x}_o$

$⟹x_o=\frac{mgsin\theta }{k}$

When the block is displaced by small unit say $x$, then it executes $SHM$ about O with an amplitude $x_o$.

Also $\overrightarrow{F_m}=q(\overrightarrow{v}\times \overrightarrow{B})$ states that the magnetic force is perpendicular to the inclined plane, hence it has no effect on $SHM$ of the block.

Net change in force experienced by the block along the inclined plane-

$F=mgsin\theta -k(x+x_o)⟹F=-kx$

$\therefore ma=-kx$

$⟹a=-\frac{k}{m}x$

Comparing with $a=-w^{2}x$

we get $w^2=\frac{k}{m}$

Time period of oscillation $T=\frac{2\pi }{w}=2\pi \sqrt{\frac{m}{k}}$