Chapter 24, Problem 8STP

To determine

To Choose: The correct option.

Answer to Problem 8STP

Option A

Explanation of Solution

Given:

Charge on proton, $q=1.6\times {10}^{-19}\text{C}$

Strength of Magnetic field, $B=0.25\text{T}$

Field is perpendicular to the motion of the charge. So, $\theta =90°$

Speed of light, $v=4.0\times {10}^6\text{m/s}$

Formula used:

Magnetic force experienced by a charge moving in a magnetic field is:

  $F=qvB\sin \theta$

Calculations:

Substitute the values:

  $\begin{array}{c}F=qvB\sin \theta \\ =1.6\times {10}^{-19}\times 0.25\times 4.0\times {10}^6\times \sin 90°\\ =1.6\times {10}^{-13}\text{N}\end{array}$

The force on the charge would be perpendicular to its motion and direction of magnetic field. Thus, the force would be directed towards left.

Conclusion:

Thus, option A is correct.