Chapter 20, Problem 60P

To determine

The magnitude and direction of electric field for the electron to follow a straight path

Answer to Problem 60P

The magnitude of electric field is $\text{1}{\text{.56×10}}^6\text{ V/m}$ and it is perpendicular to the magnetic field and initial velocity of the proton.

Explanation of Solution

Given:

The radius of circular path is

  $r\text{ = 5}\text{.10 cm }=\text{5}\text{.10}\times {\text{10}}^{-2}\text{ m}$

The magnetic field intensity B = 0.566 T $$

Formula used:

The velocity of the proton when it moves in a circular path is

  $v=\frac{eBr}{m}$$$

Charge on proton is $e=1.6\times {10}^{-19}$

Mass of proton is $m=1.673\times {10}^{-27}\text{kg}$

Calculation:

The force due to magnetic field is perpendicular to the magnetic field. Hence for a straight-line path the force due to electric field should equal and opposite to that due to magnetic field

Therefore the formula for electric field is

  $\begin{array}{l}E=vB\\ E=\frac{e{B}^{2}r}{m}\end{array}$

This electric field is perpendicular to the magnetic field and velocity .

Substituting the given values,

  $\begin{array}{c}E=\frac{\left(\text{1}{\text{.6×10}}^{\text{-19}}\text{C}\right){\left(\text{0}\text{.566T}\right)}^{\text{2}}\left(\text{5}{\text{.10×10}}^{\text{-2}}\text{m}\right)}{\left(\text{1}{\text{.673×10}}^{\text{-27}}\text{kg}\right)}\\ \text{= 1}{\text{.56×10}}^6\text{ V/m}\end{array}$$$

Conclusion:

Thus, the magnitude of electric field is

  $\text{1}{\text{.56×10}}^6\text{ V/m}$ and it is perpendicular to the magnetic field and initial velocity of the proton.

  $$