An electron is moving at a speed of $3.2 \times 10^7 \ m/s$ in a magnetic field of $12 \times 10^{-4} \ T$ perpendicular to the direction of motion of the electron. The radius of the path of the electron is . . . . . . $cm$. $(e = 1.6 \times 10^{-19} \ C$ and $m_e = 9 \times 10^{-31} \ kg)$

The radius of curvature of the path of a charged particle in a uniform magnetic field is directly proportional to

Uniform magnetic fields of different strengths ($B_1$ and $B_2$),both normal to the plane of the paper,exist as shown in the figure. $A$ charged particle of mass $m$ and charge $q$,at the interface at an instant,moves into the region $2$ with velocity $v$ and returns to the interface. It continues to move into region $1$ and finally reaches the interface. What is the displacement of the particle during this movement along the interface? (Consider the velocity of the particle to be normal to the magnetic field and $B_2 > B_1$)

The radius of the circular path of an electron when subjected to a perpendicular magnetic field is:

An $\alpha$-particle of $1 \text{ MeV}$ kinetic energy is moving on a circular path in a uniform magnetic field. The kinetic energy of a proton,to be moved in the same magnetic field on a circular path of double radius,is $..... \text{ MeV}$.

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