When an electron is moving at a speed $v$ perpendicular to a uniform magnetic field of flux density $B$, it follows a path of radius $R$ - AQA - A-Level Physics - Question 13 - 2019 - Paper 2

The radius of the path of a charged particle moving in a magnetic field is given by the formula:

$r = \frac{mv}{qB}$

Where:

• $r$ is the radius of the path,
• $m$ is the mass of the electron,
• $v$ is the velocity of the electron,
• $q$ is the charge of the electron,
• $B$ is the magnetic flux density.

For the first electron, we have: $R = \frac{mv}{qB}$

For the second electron, substituting $v$ as rac{v}{2} and $B$ as $4B$: $r_2 = \frac{m \left(\frac{v}{2}\right)}{q(4B)}$

This simplifies to: $r_2 = \frac{mv}{2qB \cdot 4} = \frac{mv}{8qB}$

From the first equation: $\frac{mv}{qB} = R$

Thus, we can express the radius for the second electron as: $r_2 = \frac{R}{8}$

Therefore, the radius of the path of the second electron is (\frac{R}{8}).