Question

An electron moves in a circular path in a region os space filled with a uniform magnetic field B= 0.4 T. to double the radius of the electron’s path, the magnitude of the magnetic field must become:
a. 0.8 T
b. 0.2 T
c. 0.1 T
d. 0.3 T
e. zero

Answer 1

Answer:

$0.2\; {\rm T}$, assuming that the speed of the electron stays the same.

Explanation:

Let $v$ denote the speed of this electron. Let $q$ denote the electric charge on this electron. Let $m$ denote the mass of this electron.

Since the path of this electron is a circle (not a helix,) this path would be in a plane normal to the magnetic field.

Let $B$ denote the strength of this magnetic field. The size of the magnetic force on this electron would be:

$F = q\, v\, B$.

Assuming that there is no other force on this electron. The net force on this electron would be $F = q\, v\, B$. By Newton's Second Law of motion, the acceleration of this electron would be:

$\begin{aligned}a &= \frac{F}{m} \\ &= \frac{q\, v\, B}{m}\end{aligned}$.

On the other hand, since this electron is in a circular motion with a constant speed:

$\begin{aligned} a = \frac{v^{2}}{r} \end{aligned}$.

Combine the two equations to obtain a relationship between $r$ (radius of the path of the electron) and $B$ (strength of the magnetic field:)

$\begin{aligned}\frac{q\, v\, B}{m} = \frac{v^{2}}{r}\end{aligned}$.

Simplify to obtain:

$\begin{aligned}r &= \frac{m\, v^{2}}{q\, v\, B} \\ &= \frac{m\, v}{q\, B} \\ &= \left(\frac{m\, v}{q}\right)\, \frac{1}{B}\end{aligned}$.

In other words, if the speed $v$ of this electron stays the same, the radius $r$ of the path of this electron would be inversely proportional to the strength $B$ of the magnetic field. Doubling the radius of this path would require halving the strength of the magnetic field (to $0.2\; {\rm T}$.)