A proton enters a magnetic field of flux density $1.5\,weber/{m^2}$ with a velocity of $2 \times {10^7}\,m/\sec $ at an angle of $30^\circ $ with the field. The force on the proton will be
$2.4 \times {10^{ - 12}}\,N$
$0.24 \times {10^{ - 12}}\,N$
$24 \times {10^{ - 12}}\,N$
$0.024 \times {10^{ - 12}}\,N$
The motion of a charged particle can be used to distinguish between a magnetic field and electric field in a certain region by firing the charge
Two particles of charges $+Q$ and $-Q$ are projected from the same point with a velocity $v$ in a region of uniform magnetic field $B$ such that the velocity vector makes an angle $q$ with the magnetic field. Their masses are $M$ and $2M,$ respectively. Then, they will meet again for the first time at a point whose distance from the point of projection is
A proton (mass $m$ ) accelerated by a potential difference $V$ flies through a uniform transverse magnetic field $B.$ The field occupies a region of space by width $'d'$. If $\alpha $ be the angle of deviation of proton from initial direction of motion (see figure), the value of $sin\,\alpha $ will be
A proton (mass $ = 1.67 \times {10^{ - 27}}\,kg$ and charge $ = 1.6 \times {10^{ - 19}}\,C)$ enters perpendicular to a magnetic field of intensity $2$ $weber/{m^2}$ with a velocity $3.4 \times {10^7}\,m/\sec $. The acceleration of the proton should be
A proton and an alpha particle are separately projected in a region where a uniform magnetic field exists. Their initial velocities are perpendicular to direction of magnetic field. If both the particles move around magnetic field in circles of equal radii, the ratio of momentum of proton to alpha particle $\left( {\frac{{{P_p}}}{{{P_\alpha }}}} \right)$ is