Chapter 21, Problem 1P

In New England, the horizontal component of the earth's magnetic field has a magnitude of $1.6\times {10}^{-5} T$. An electron is shot vertically straight up from the ground with a speed of $2.1\times {10}^6 m/s$. What is the magnitude of the acceleration caused by the magnetic force? Ignore the gravitational force acting on the electron.

To determine

The acceleration of the electron.

Answer to Problem 1P

  $5.907\times {10}^{12}m/s^2$

Explanation of Solution

Given:

Magnitude of magnetic field: $1.6\times {10}^{-5}T$

Initial velocity of the electron: $2.1\times {10}^{6}m/s$

Formula used:

To calculate the force exerted by the magnetic field, Use the formula:

  $F=q\left(V\times B\right)$

Where, q is the charge, V is the velocity and B is the magnetic field.

And to calculate the acceleration use the formula:

  $F=ma$

Where, m is the mass and a is the acceleration.

Calculation:

Substituting the known values in the formula for force:

  $\begin{array}{l}\Rightarrow F=q\left(V\times B\right)\\ \Rightarrow F=\left(1.6\times {10}^{-19}\right)\left(2.1\times {10}^6\times 1.6\times 10^{-5}\right)\\ \Rightarrow F=5.376\times {10}^{-18}N\end{array}$

Now plug the value of force and mass of electron in the equation for acceleration.

  $\begin{array}{l}\Rightarrow F=ma\\ \Rightarrow a=\frac{F}{m}\\ \Rightarrow a=\frac{5.376\times {10}^{-18}}{9.1\times {10}^{-31}}\\ \Rightarrow a=5.907\times {10}^{12}m/s^2\end{array}$

Conclusion:

Thus, the magnitude of acceleration of the electron is: $5.907\times {10}^{12}m/s^2$