Chapter 24.2, Problem 26PP

To determine

Magnitude and direction of force on proton.

Answer to Problem 26PP

  $F=3.2\times {10}^{-12}\text{N}$ and the direction is upwards.

Explanation of Solution

Given:

The given figure is shown below.

  

Formula Used:

The force on a moving charge (Lorentz force) is,

  $F=qvB\sin \theta$

Where, $q$ is the moving charge, $v$ is velocity, $B$ is magnetic field strength and $\theta$ is the minimum angle between the direction of $v\text{and }B$ .

Calculation:

From the given figure,

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

Velocity, $v=4\times {10}^7\text{m/s}$

Strength of magnetic field, $B=0.5\text{T}$

Angle between $v\text{and }B$ is $\theta =90°$

So, on putting the values,

  $\begin{array}{c}F=\left(+1.6\times {10}^{-19}\right)\times \left(4\times {10}^7\right)\times 0.5\times \sin 90°\\ =3.2\times {10}^{-12}\text{N}\end{array}$

Now, the direction of force is given by right hand rule.

Since the velocity is in right direction and direction of magnetic field is inward so, by right hand rule the direction of force is up.

Conclusion:

Therefore, the force on the proton is $F=3.2\times {10}^{-12}\text{N}$ and the direction is upwards.