91Ó°ÊÓ

The Lorentz force acting on a charged particle moving in a magnetic field is given by the following formula: $$ F = q \cdot v \cdot B $$ Where \(F\) is the magnitude of the Lorentz force, \(q\) is the charge of the particle, \(v\) is the particle's speed, and \(B\) is the magnetic field strength.

In circular motion, the centripetal force is given by the formula: $$ F_c = \frac{m \cdot v^2}{r} $$ Where \(F_c\) is the magnitude of the centripetal force, \(m\) is the mass of the particle, and \(r\) is the radius of the circular path. Since the Lorentz force acts as the centripetal force in this case, we have: $$ q \cdot v \cdot B = \frac{m \cdot v^2}{r} $$

Solve for r in the equation found in Step 2: $$ r = \frac{m \cdot v}{q \cdot B} $$

The distance traveled by the charged particle in one full revolution is the circumference of the circular path, which is given by the formula: $$ C = 2 \pi r $$ The time taken for one complete revolution, or the period of the motion (\(T\)), is the distance traveled divided by the speed: $$ T = \frac{C}{v} = \frac{2 \pi r}{v} $$

Insert the expression for \(r\) found in Step 3 into the equation for \(T\) found in Step 4: $$ T = \frac{2 \pi \cdot \frac{m \cdot v}{q \cdot B}}{v} $$ Simplify the expression by canceling the \(v\) term, as our goal is to show that the period is independent of the particle's speed. The final expression for the period is: $$ T = \frac{2 \pi m}{q \cdot B} $$ This expression shows that the period (\(T\)) for one revolution of a charged particle moving perpendicular to a uniform magnetic field is independent of the particle's speed, as it does not contain any term related to the speed \(v\). The period is determined solely by the mass of the particle (\(m\)), charge of the particle (\(q\)), and the magnetic field strength (\(B\)).