Chapter 25, Problem 100A

To determine

To calculate:

The force on the electron and the acceleration of the electron.

Answer to Problem 100A

The force on the electron is $2.7\times {10}^{-13}$ N an acceleration of the electron is $3.0\times {10}^{17}$ m/s².

Explanation of Solution

Given:

For an electron,

Velocity: $2.1\times {10}^6$ m/s

Magnetic Field: $0.81$ T

Mass: $9.11\times {10}^{-31}$ kg

Formula Used:

Formula used for force when charge and magnetic field is given,

  $F=qvB$

Where,

  $F$ : Force

  $q$ : Charge

  $v$ : Velocity

  $B$ : Magnetic Field

And

  $F=ma$

Where,

  $F$ : Force

  $m$ : Mass

  $a$ : Acceleration

Calculation:

First calculate the force on an electronthe formula of force when the charge and magnetic field is given,

  $F=qvB$

Consider the charge on electron as $1.60\times {10}^{-19}C$

Substitute all the given values in the above formula,

  $F=qvB$

  $=$$\left(\text{1}{\text{.60×10}}^{\text{-19}}\text{C}\right)\text{×}\left(\text{2}{\text{.1×10}}^{\text{6}}\text{m/s}\right)\text{×}\left(\text{0}\text{.81}\text{T}\right)$

  $=$$\left(3.36\times {10}^{-13}\right)\times \left(0.81\right)$

  $F=$$2.7\times {10}^{-13}$ N

Now, in the second step calculate the acceleration of an electron when the mass and force is given,

  $F=ma$

Rearrange the formula to find acceleration,

  $a=\frac{F}{m}$

Substitute all the given values in the above formula,

  $\frac{\text{2}{\text{.7×10}}^{\text{-13}}\text{N}}{\text{9}{\text{.11×10}}^{\text{-31}}\text{kg}}$

  $a$$=$$3.0\times {10}^{17}{\text{m/s}}^{\text{2}}$

Conclusion:

Hence, force on the electron is $2.7\times {10}^{-13}\text{N}$ and acceleration of the electron is $3.0\times {10}^{\text{17}}$ m/s²

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