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)
$2\pi \sqrt {\frac{m}{k}}$
$2\pi \sqrt {\frac{{2m}}{k}}$
$2\pi \sqrt {\frac{{qB}}{K}}$
$2\pi \sqrt {\frac{{qB}}{{2K}}}$
An electron moves along vertical line and away from the observer, then pattern of concentric circular magnetic field lines which are produced due to its motion
A particle of mass $m = 1.67 \times 10^{-27}\, kg$ and charge $q = 1.6 \times 10^{-19} \, C$ enters a region of uniform magnetic field of strength $1$ $tesla$ along the direction shown in the figure. the radius of the circular portion of the path is :-
A proton moving with a constant velocity, passes through a region of space without change in its velocity. If $E$ $\& B$ represent the electric and magnetic fields respectively, this region may have
A negative charge is coming towards the observer. The direction of the magnetic field produced by it will be (as seen by observer)
A proton is projected with a velocity $10^7\, m/s$, at right angles to a uniform magnetic field of induction $100\, mT$. The time (in second) taken by the proton to traverse $90^o$ arc is $(m_p = 1.65\times10^{-27}\, kg$ and $q_p = 1.6\times10^{-19}\, C)$