If a particle of charge ${10^{ - 12}}\,coulomb$ moving along the $\hat x - $ direction with a velocity ${10^5}\,m/s$ experiences a force of ${10^{ - 10}}\,newton$ in $\hat y - $ direction due to magnetic field, then the minimum magnetic field is
$6.25 \times {10^3}\,tesla$ in $\hat z - $ direction
${10^{ - 15}}\,tesla$ in $\hat z - $ direction
$6.25 \times {10^{ - 3}}\,tesla$ in $\hat z - $ direction
${10^{ - 3}}\,tesla$ in $\hat z - $ direction
When a charged particle moving with velocity $\vec V$ is subjected to a magnetic field of induction $\vec B$ , the force on it is non-zero. This implies the
A particle of charge $q$, mass $m$ enters in a region of magnetic field $B$ with velocity $V_0 \widehat i$. Find the value of $d$ if the particle emerges from the region of magnetic field at an angle $30^o$ to its ititial velocity:-
A proton and a deutron both having the same kinetic energy, enter perpendicularly into a uniform magnetic field $B$. For motion of proton and deutron on circular path of radius ${R_p}$ and ${R_d}$ respectively, the correct statement is
An electron, moving in a uniform magnetic field of induction of intensity $\vec B,$ has its radius directly proportional to
A particle with charge $-Q$ and mass m enters a magnetic field of magnitude $B,$ existing only to the right of the boundary $YZ$. The direction of the motion of the $m$ particle is perpendicular to the direction of $B.$ Let $T = 2\pi\frac{m}{{QB}}$ . The time spent by the particle in the field will be