When an electron moves at a speed $v$ perpendicular to a uniform magnetic field of flux density $B$, the radius of its path is $R$ - AQA - A-Level Physics - Question 14 - 2022 - Paper 2

To find the radius of the path of the second electron, we can use the formula for the radius of curvature in a magnetic field:

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

For the first electron:

• The speed is $v$.
• The magnetic flux density is $B$.
• The radius is $R$, so:

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

For the second electron:

• The speed is rac{v}{2}.
• The magnetic flux density is $4B$.

Let the radius for the second electron be $r$. We have:

$r = \frac{m(\frac{v}{2})}{q(4B)}$

This can be simplified to:

$r = \frac{mv/2}{4qB} = \frac{mv}{8qB}$

We know from the first electron that:

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

Thus, we can express $r$ in terms of $R$:

$r = \frac{R}{8}$

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