Chapter 19, Problem 117P

(a)

To determine

The electromagnetic force on an electron.

Answer to Problem 117P

The electromagnetic force on an electron is $\underset{\_}{8.6\times {10}^{-15}\text{N at 68}°\text{ below west}}$.

Explanation of Solution

Figure 1 gives the direction of charge with helicopter view looking down. The cross product of peed and magnetic field is up and perpendicular to the plane.

Write the equation of magnitude of the resultant electromagnetic force.

  $F=\sqrt{F_E^2+F_B^2}$                                                                                                    (I)

Here, $F$ is the net electromagnetic force, $F_E$ is the force due to electric field and $F_B$ is the force due to magnetic field.

Write the equation of angle of the resultant electromagnetic force.

  $\theta ={\tan }^{-1}\left(\frac{F_B}{F_E}\right)$                                                                                                  (II)

Here, $\theta$ is the angle.

Write the equation for the force due to electric field.

  $F_E=eE$                                                                                                           (III)

Here, $e$ is the electronic charge and $E$ is the electric field.

Write the equation for the force due to magnetic field.

  $F_B=evB$                                                                                                           (IV)

Here, $v$ is the velocity and $B$ is the magnetic field.

Rewrite the expression from equation (I).

  $F=\sqrt{{\left(eE\right)}^2+{\left(evB\right)}^2}$                                                                                        (V)

Rewrite the expression from equation (II).

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{\left(evB\right)}{\left(eE\right)}\right)\\ ={\tan }^{-1}\left(\frac{\left(vB\right)}{\left(E\right)}\right)\end{array}$                                                                                          (VI)

Conclusion:

Substitute $1.602\times {10}^{-19}\text{C}$ for $e$, $2.0\times {10}^4\text{V/m}$ for $E$, $0.0050\text{T}$ for $B$ and $1.0\times {10}^7\text{m/s}$ for $v$ in equation (V) to find $F$.

  $\begin{array}{c}F=\sqrt{{\left[\left(1.602\times {10}^{-19}\text{C}\right)\left(2.0\times {10}^4\text{V/m}\right)\right]}^2+{\left[\left(1.602\times {10}^{-19}\text{C}\right)\left(1.0\times {10}^7\text{m/s}\right)\left(0.0050\text{T}\right)\right]}^2}\\ =8.6\times {10}^{-15}\text{N}\end{array}$

Substitute, $2.0\times {10}^4\text{V/m}$ for $E$, $0.0050\text{T}$ for $B$ and $1.0\times {10}^7\text{m/s}$ for $v$ in equation (VI) to find $\theta$.

  $\begin{array}{c}\theta =\frac{\left[\left(1.0\times {10}^7\text{m/s}\right)\left(0.0050\text{T}\right)\right]}{\left[\left(2.0\times {10}^4\text{V/m}\right)\right]}\\ =68°\end{array}$

Thus, the electromagnetic force on an electron is $\underset{\_}{8.6\times {10}^{-15}\text{N at 68}°\text{ below west}}$.

(b)

To determine

There is any velocity which make the electromagnetic force will be zero.

Answer to Problem 117P

No such velocity is there which will produce zero electromagnetic force.

Explanation of Solution

The electromagnetic force will not be zero. The electric force and magnetic force is perpendicular to each other and having different magnitude. If there is change in velocity also, the net electromagnetic force will not be zero.

To make the electromagnetic force zero, the force due to electric and magnetic field should be equal and opposite direction. Since the magnitude is different and the direction is perpendicular to each other, the electromagnetic force will not be zero. Thus, any change in velocity cannot make the electromagnetic force zero.