Question

question 8 of 12 current attempt in progress what uniform magnetic field, applied perpendicular to a beam of electrons moving at 3.01×10^6 m/s, is required to make the electrons travel in a circular arc of radius 0.320 m? units attempts: 0 of 3 used save for later etextbook and media view policies submit answer - /0.5

Explanation:

Step1: Recall the formula for the radius of the circular path of a charged - particle in a magnetic field

The formula is $r=\frac{mv}{qB}$, where $r$ is the radius of the circular path, $m = 9.11\times10^{- 31}\text{ kg}$ is the mass of an electron, $v$ is the velocity of the electron, $q=1.6\times10^{-19}\text{ C}$ is the charge of an electron, and $B$ is the magnetic field strength. We need to solve for $B$, so we can re - arrange the formula to $B=\frac{mv}{qr}$.

Step2: Substitute the given values

We are given that $r = 0.320\text{ m}$, $v=3.01\times10^{6}\text{ m/s}$, $m = 9.11\times10^{-31}\text{ kg}$, and $q = 1.6\times10^{-19}\text{ C}$.
\[

$$\begin{align*} B&=\frac{(9.11\times 10^{-31}\text{ kg})\times(3.01\times 10^{6}\text{ m/s})}{(1.6\times 10^{-19}\text{ C})\times(0.320\text{ m})}\\ &=\frac{9.11\times3.01\times10^{-31 + 6}}{1.6\times0.320\times10^{-19}}\text{ T}\\ &=\frac{27.4211\times10^{-25}}{0.512\times10^{-19}}\text{ T}\\ &=\frac{27.4211}{0.512}\times10^{-25 + 19}\text{ T}\\ &= 53.56\times10^{-6}\text{ T}\\ &=5.36\times 10^{-5}\text{ T} \end{align*}$$

\]

Answer:

$5.36\times 10^{-5}\text{ T}$