Question

A proton of mass m and charge q enters a magnetic field B perpendicularly with speed v. What is the time period of its circular motion?

Answer 1

$\frac{2\pi m}{qB}$

Answer 2

When a charged particle enters a magnetic field perpendicularly, the magnetic force acts as the centripetal force, causing the particle to move in a circular path.

1. Magnetic Force: The force experienced by a charged particle $q$ moving with velocity $v$ in a magnetic field $B$ is given by $\vec{F_B} = q(\vec{v} \times \vec{B})$. Since the proton enters perpendicularly, the angle between $\vec{v}$ and $\vec{B}$ is $90^\circ$, so the magnitude of the magnetic force is: $F_B = qvB \sin(90^\circ) = qvB$

2. Centripetal Force: For a particle of mass $m$ moving in a circle of radius $r$ with speed $v$, the centripetal force required is: $F_c = \frac{mv^2}{r}$

3. Equating Forces: Since the magnetic force provides the necessary centripetal force for circular motion: $F_B = F_c$ $qvB = \frac{mv^2}{r}$

4. Radius of the Circular Path: From the above equation, we can find the radius $r$: $r = \frac{mv^2}{qvB} = \frac{mv}{qB}$

5. Time Period of Circular Motion: The time period $T$ is the time taken to complete one full revolution. It is given by the circumference of the circle divided by the speed: $T = \frac{2\pi r}{v}$

6. Substituting the Radius: Substitute the expression for $r$ into the equation for $T$: $T = \frac{2\pi}{v} \left( \frac{mv}{qB} \right)$ $T = \frac{2\pi m}{qB}$

The time period of the proton's circular motion is $\frac{2\pi m}{qB}$.