Animated Solution for Physics - Magnetic Effects of Current: In an experiment, electrons are accelerated, from rest by applying a voltage of $500\text{ V}$. Calculate the radius of the path, if a magnetic field $100\text{ mT}$ is then applied. (Take, charge of the electron $=1.6\times 10^{-19}\text{ C}$ and mass of the electron $=9.1\times 10^{-31}\text{ kg}$)

Imagine an electron, initially at rest, waiting to begin its journey. We kickstart its motion by applying a potential difference of $500\text{ V}$. This electric field acts as a slingshot, accelerating the electron and giving it a substantial amount of kinetic energy.

As soon as it leaves the accelerating region, it enters a new domain: a uniform magnetic field of $100\text{ mT}$ directed perpendicular to its velocity. According to the Lorentz force law, a magnetic field exerts a force perpendicular to both the velocity and the field itself. This continuous perpendicular force acts as a centripetal force, forcing the electron to move in a perfect circular path.

To find the radius of this circular path, we need to connect the energy gained to the dynamics of circular motion. First, the kinetic energy $K$ gained by the electron is equal to the work done by the electric field:

We also know that kinetic energy is intimately related to momentum $p$:

Substituting the kinetic energy, the momentum of the electron as it enters the magnetic field is:

Now, for a charged particle moving in a circular path within a magnetic field, the magnetic force provides the necessary centripetal force ($evB=\frac{m{v}^2}{r}$). Rearranging this gives us the radius $r$:

Combining our equations, we get the master formula for the radius:

Now comes the crucial part—substituting the values without making any silly mistakes. We are given: - Mass of electron, $m=9.1\times 10^{-31}\text{ kg}$ - Charge of electron, $e=1.6\times 10^{-19}\text{ C}$ - Accelerating voltage, $\Delta V=500\text{ V}$ - Magnetic field, $B=100\text{ mT}=100\times 10^{-3}\text{ T}=10^{-1}\text{ T}$

Plugging these into our master equation:

Let's simplify the terms inside the square root. Grouping the numbers and the powers of 10:

Taking the square root ($\sqrt{145.6}\approx 12.06$ and $\sqrt{10^{-48}}=10^{-24}$):

Dividing the numbers:

Rounding to the given options, the radius of the path is $7.5\times 10^{-4}\text{ m}$. The elegance of physics lies in how perfectly energy and forces balance out to dictate the exact trajectory of a fundamental particle!