Chapter 5, Problem 1P

An electron with a speed of 8 × 10⁶ m/s is projected along the positive x direction into a medium containing a uniform magnetic flux density $B=(\hat{x}4-\hat{z}3)T$. Given that e = 1.6 × 10⁻¹⁹ C and the mass of an electron is m_e = 9.1 × 10⁻³¹ kg, determine the initial acceleration vector of the electron (at the moment it is projected into the medium).

To determine

The initial acceleration vector of the electron.

Answer to Problem 1P

The initial acceleration vector of the electron is $\underset{\_}{-\widehat{y}\left(4.22\times {10}^{18}\right)\text{m}/{\text{s}}^{\text{2}}}$.

Explanation of Solution

Given data:

The speed of electron in the positive $x$ direction is $8\times {10}^6\text{m}/\text{s}$.

The magnetic flux density is $\left(\widehat{x}4-\widehat{z}3\right)T$.

The charge of electron is $1.6\times {10}^{-19}\text{C}$.

The mass of an electron is $9.1\times {10}^{-31}\text{kg}$.

Calculation:

The cross product is given as,

$\begin{array}{c}u\times B=\left|\begin{array}{ccc}\widehat{x}& \widehat{y}& \widehat{z}\\ 8\times {10}^6& 0& 0\\ 4& 0& -3\end{array}\right|\\ =\widehat{x}\left(0-0\right)-\widehat{y}\left(-24\times {10}^6-0\right)+\widehat{z}\left(0-0\right)\\ =\widehat{y}\left(24\times {10}^6\right)\end{array}$

Here,

$u$ is the speed of the electron in the positive $x$ direction and

$B$ is the magnetic flux density.

The acceleration vector of the particle is given as,

$a=-\frac{e}{m_{\text{e}}}\left(u\times B\right)$

Here,

$a$ is the acceleration vector of the particle.

$e$ is the charge of an electron.

$m_{\text{e}}$ is mass of an electron.

Substitute $1.6\times {10}^{-19}$ for $e$, $9.1\times {10}^{-31}$ for $m_{\text{e}}$ and $\widehat{y}\left(24\times {10}^6\right)$ for $\left(u\times B\right)$ in the above formula.

$\begin{array}{c}a=-\frac{1.6\times {10}^{-19}}{9.1\times {10}^{-31}}\left(24\times {10}^6\widehat{y}\right)\\ =-\widehat{y}\left(4.22\times {10}^{18}\right)\text{m}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the initial acceleration vector of the electron is $\underset{\_}{-\widehat{y}\left(4.22\times {10}^{18}\right)\text{m}/{\text{s}}^{\text{2}}}$.