Chapter 30, Problem 14SP

A proton travels at 100 m/s in and parallel to a 2.00-T magnetic field. What is the magnitude of the force on the proton?

To determine

The magnitude of the force on the proton, which travels at a speed of $100\text{ m/s}$ in and parallel to a $2.00\text{ T}$ magnetic field.

Answer to Problem 14SP

Solution:

$0\text{ N}$ or no force

Explanation of Solution

Given data:

The velocity of the proton is $100\text{ m/s}$.

Themagnitude of the magnetic field acting on the proton is $2.00\text{ T}$.

Formula used:

The expression for a force acting on a charge moving in a magnetic field is written as

$F_M=qvB\sin \theta$

Here, $F_M$ is the force acting on a charge due to the presence of a magnetic field, $q$ is the amount of charge, $v$ is the speed of the charge, $B$ is the strength of the magnetic field, and $\theta$ is the angle between the velocity of the charge and the direction of the magnetic field.

Explanation:

Recall the expression fora force acting on a charge moving in a magnetic field:

$F_M=qvB\sin \theta$

Since the charge is travelling parallel to the magnetic field, the angle between the velocity of the charge and the direction of the magnetic field becomes $0°$.

Substitute $1.6\times {10}^{-19}\text{ C}$ for $q$, $100\text{ m/s}$ for $v$, $2.00\text{ T}$ for $B$, and $0°$ for $\theta$

$\begin{array}{c}{F}_M=\left(1.6\times {10}^{-19}\text{ C}\right)\left(100\text{ m/s}\right)\left(2.00\text{ T}\right)\sin \left(0°\right)\\ =0\text{ N}\end{array}$

The charge on the proton is $1.6\times {10}^{-19}\text{ C}$. Understand that the direction of the velocity of the proton and the magnetic field are in the same direction. So, the angle between them will be $0$.

Conclusion:

Hence, the magnitude of the force acting on the proton is $0\text{ N}$.