An electron (mass $= 9 \times 10^{-31} \, kg$,charge $= 1.6 \times 10^{-19} \, C$) with a kinetic energy of $7.2 \times 10^{-18} \, J$ is moving in a circular orbit in a magnetic field of $9 \times 10^{-5} \, Wb/m^2$. The radius of the orbit is ..... $cm$.

$A$ charged particle enters a magnetic field $B$ with its initial velocity making an angle of $45^\circ$ with $B$. The path of the particle will be

$A$ proton with kinetic energy of $1 \; MeV$ moves from south to north. It experiences an acceleration of $10^{12} \; m/s^2$ due to an applied magnetic field (directed from west to east). The value of the magnetic field is: ....... $mT$ (Rest mass of proton is $1.6 \times 10^{-27} \; kg$)

$A$ charged particle moving with a velocity $\vec{v} = v_1 \hat{i} + v_2 \hat{j}$ in a magnetic field $\vec{B}$ experiences a force $\vec{F} = F_1 \hat{i} + F_2 \hat{j}$. Here $v_1, v_2, F_1, F_2$ are all constants. Then $\vec{B}$ can be

$A$ small block of mass $20 \,g$ and charge $4 \,mC$ is released on a long smooth inclined plane of inclination angle $45^{\circ}$. $A$ uniform horizontal magnetic field of $1 \,T$ is acting parallel to the surface, as shown in the figure. The time from the start when the block loses contact with the surface of the plane is (in $\,s$)

Derive the expression for the magnetic force acting on a moving charge $q$ with velocity $\vec{v}$ in a uniform magnetic field $\vec{B}$.