A particle of mass m and positive charge q travels with a uniform...

When a particle of mass $m$ and positive charge $q$ enters a uniform magnetic field $\vec{B}$ perpendicular to its velocity, it moves in a circular path with an angular speed $\omega = \dfrac{qB}{m}$.

The magnetic field is directed into the page. Applying the right-hand rule for the magnetic force $\vec{F} = q(\vec{v} \times \vec{B})$, the force on the positive charge at the point of entry is directed upwards and to the right. This indicates that the center of the circular path lies outside the magnetic field region.

The particle enters the field at an angle $\theta$ with the normal. Due to the symmetry of the circular path, it will exit the field at the same angle $\theta$ with the normal.

The angle that the velocity vector turns through is equal to the angle subtended by the circular arc at the center. From the geometry of the path, the radii at the entry and exit points each make an angle $\theta$ with the boundary. The angle subtended by the arc at the center is $\phi = \pi - 2\theta$.

The time spent by the particle inside the magnetic field is given by $t = \dfrac{\phi}{\omega}$.

Substituting the values, we get $t = \dfrac{\pi - 2\theta}{\dfrac{qB}{m}} = \dfrac{m}{qB}(\pi - 2\theta)$.

Answer: $\dfrac{m}{q B}(\pi - 2\theta)$