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A centripetal force is the inward resultant force needed for circular motion. Newton's second law and centripetal acceleration give magnitude $m{v}^2/r$ for a particle of mass $m$, speed $v$, and path radius $r$. It is a role played by an actual interaction, such as tension or an electric force, rather than an additional interaction to add to a free-body diagram. A purely radial centripetal force is perpendicular to the velocity, so it does no work and leaves the speed unchanged.

Put $\mu =GM$. Balancing Newton's law of universal gravitation with the required centripetal force gives $m{v}_0^2/r_0=\mu m/r_0^2$, henceThe circular Kepler orbit is stable against small radial perturbations: its effective radial potential is $U_{\mathrm{eff}}=J^2/(2m{r}^2)-\mu m/r$, and at the circular value $J^2=\mu m^2{r}_0$, $U_{\mathrm{eff}}^{\prime \prime }(r_0)=\mu m/r_0^3>0$. Here $J$ is the conserved angular momentum.

Let ${\mathbf{r}}_P$ be the initial position, $\hat{\mathbf{b}}=\mathbf{B}/B$, and $\Omega =qB/m$, including the sign of $q$. Solving $\dot{\mathbf{v}}=\Omega \mathbf{v}\times \hat{\mathbf{b}}$ givesThusThe locus is a circle in the plane perpendicular to $\mathbf{B}$, centred at ${\mathbf{r}}_c={\mathbf{r}}_P+({\mathbf{v}}_0\times \hat{\mathbf{b}})/\Omega$, withThis also follows by equating the magnetic force to the centripetal force. The nonrelativistic angular frequency is $|q|B/m$.

Take the initial velocity perpendicular to a uniform magnetic field. The magnetic Lorentz force is perpendicular to the motion, has constant magnitude $|q|vB$, and supplies the centripetal force:The rotation sense is fixed by $q\mathbf{v}\times \mathbf{B}$. A component of velocity parallel to the field remains constant and produces a helical motion in a uniform magnetic field; a circle requires that component to vanish. The speed-independent cyclotron frequency here assumes nonrelativistic motion.

Choose the entry point as $(0,0)$, the initial velocity as $v\hat{\mathbf{x}}$, and the slit as $(d,a)$, with $a,d>0$. A magnetic field perpendicular to the page changes the direction of the velocity, while the Lorentz force does no work. For positive $q$, an upward initial force requires the magnetic flux density to point into the page. The resulting circular motion has centre $(0,r)$, since its initial tangent is horizontal.

The slit must lie on the circle, giving $d^2+(a-r{)}^2=r^2$ and hence $r=(d^2+a^2)/(2a)$. Equating the centripetal force to the magnetic part of the Lorentz force, $m{v}^2/r=|q|vB$, givesA negative electric charge requires the opposite direction of magnetic flux density. These expressions assume nonrelativistic motion and negligible energy loss through electromagnetic radiation.

For an opaque screen, reaching the slit must be the first encounter with $x=d$. The velocity there has $v_x/v=(d^2-a^2)/(d^2+a^2)$, so passage from left to right requires $a<d$, as in the drawn arrangement. At $a=d$ the trajectory merely touches the screen. If $a>d$, the same circle first meets the screen at height $d^2/a<a$, where the particle is blocked; its later passage through $(d,a)$ does not solve the opaque-screen problem. This qualification distinguishes a geometric circle through two points from an accessible slit trajectory.

The electron follows a semicircle under the transverse Lorentz force. Parallel, oppositely directed incoming and outgoing paths are separated by a diameter, so the orbit radius is $r=1.00\mathrm{cm}$. Acceleration from rest through the electric potential difference $V$ gives kinetic energy $m_e{v}^2/2=eV$, where $e>0$ is the elementary electric charge. Equating the Lorentz force magnitude to the required centripetal force givesThe magnetic field must be perpendicular to the plane of the turn, with its sign chosen for the electron's negative electric charge. In the illustration the electron enters rightwards at the upper endpoint, so a magnetic field into the page bends it downwards.

The nonrelativistic estimate has $v\simeq 0.198c$. If the small special relativity correction is retained, use $pc=\sqrt{K(K+2m_e{c}^2)}$ with $K=eV$ and $B=p/(er)$. This gives $B\simeq 3.389\times 10^{-2}\mathrm{T}$, about $0.49\%$ larger. The leading answer is adequate at the precision implicit in the classical treatment.

Neglect the entrance and exit capacitor fringing fields and use the coaxial electric field between the cylindrical plates. Let $U=V(R_1)-V(R_2)$, with $R_2>R_1$. The radial electric field isFor an electron, of electric charge $-e$, to follow the circular path $R=(R_1+R_2)/2$, its inward centripetal force must be $m_e{v}_0^2/R$. ThereforeThe inner plate must be at higher electric potential: the electric field points outward and the electron's electric force inward. The electric force is perpendicular to the velocity, so no work is done and the speed remains $v_0$. The required voltage is independent of the opening angle $\varphi$, which only sets the transit time $R\varphi /v_0$. For a narrow gap, $\ln (R_2/R_1)\simeq (R_2-R_1)/R$. This is a nonrelativistic, negligible-capacitor fringing field answer; finite open cylindrical sections do not have an exactly radial electric field at their ends.

Let $U>0$ be the drop in electric potential along the positive electric charge's path. By conservation of energy, $qU=m{v}^2/2$, so the nonrelativistic exit speed is $v=\sqrt{2qU/m}$. Apply a uniform magnetic field perpendicular to the incident velocity and to the desired orbital plane. The magnetic Lorentz force is perpendicular to the velocity, does no work, and supplies the centripetal force:For example, $\mathbf{v}=v\hat{\mathbf{x}}$ and $\mathbf{B}=B\hat{\mathbf{z}}$ give initial deflection along $-y$ for positive $q$. Reversing the magnetic field reverses the bending. A uniform electric field alone would give constant acceleration and a parabolic path, not uniform circular motion. The displayed result assumes $qU\ll m{c}^2$; the exact relativistic momentum relation would give $R=\sqrt{q^2{U}^2/c^2+2mqU}/(qB)$.