Chapter 11, Problem 11.1CYU

Check Your Understanding Repeat the previous problem with the magnetic field in the x-direction rather than in the z-direction. Check your answers with RHR-1.

To determine

(a)

To Calculate:

The magnitude of the force on the alpha particle when it is moving in the positive x-direction in a magnetic field parallel to the positive x-axis.

Answer to Problem 11.1CYU

The magnetic force on the alpha particle is zero.

Explanation of Solution

Given:

Charge on the alpha particle, $q=3.2\times {10}^{-19}C$

Magnetic field parallel to the positive x-axis, $B=1.5\text{ T}$

Velocity of the alpha particle in the x-direction, $v=5\times {10}^4\text{ m/s}$

Formula used:

The magnetic force on a charged particle is given as

  ${\overrightarrow{F}}_B=q(\overrightarrow{v}\times \overrightarrow{B})$ or $F_B=qvB\sin \theta$ .-----(1)

Here $q$ is the charge of the particle, $v$ is the velocity of the particle, $B$ is the magnetic field

& $\theta$ is the angle between the velocity vector and the magnetic field vector.

Calculation:

The magnetic force on the alpha particle is given by

  $F_B=qvB\sin \theta$

But note $\theta$ is the angle between velocity of the particle and the direction of the applied magnetic field is 0 here as both are in the x-direction. Since $\theta$ is zero therefore from equation (1) the force also must be zero as $\text{sin0=0}$.

To determine

(b)

To Calculate:

The magnitude of the force on the alpha particle when it is moving in the negative y-direction in a magnetic field parallel to the positive x-axis.

Answer to Problem 11.1CYU

The magnetic force on the alpha particle is $2.4\times {10}^{-14}$ N towards positive z-axis.

Explanation of Solution

Given:

Charge on the alpha particle, $q=3.2\times {10}^{-19}C$

Magnetic field parallel to the positive x-axis, $B=1.5\text{ T}$

Velocity of the alpha particle in the negative y-direction, $\overrightarrow{v}=-5\times {10}^4\text{ <mover accent="true"><mo>j</mo><mo>⌢</mo></mover><mo> m/s</mo>}$

Formula used:

The magnetic force on a charged particle is given as

  ${\overrightarrow{F}}_B=q(\overrightarrow{v}\times \overrightarrow{B})$ or $F_B=qvB\sin \theta$ .-----(1)

Here $q$ is the charge of the particle, $v$ is the velocity of the particle, $B$ is the magnetic field

& $\theta$ is the angle between the velocity vector and the magnetic field vector.

Calculation:

The magnetic force on the alpha particle is given by

  $\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$

Substituting the value of $q$, $\overrightarrow{v}$ & $\overrightarrow{B}$ we get,

  $\overrightarrow{F}=(3.2\times {10}^{-19})\left[(-5\times {10}^4\text{ <mover accent="true"><mo>j</mo><mo>⌢</mo></mover>})\times (1.5\stackrel{⌢}{i})\right]$

Or $\overrightarrow{F}=2.4\times {10}^{-14}\stackrel{⌢}{k}\text{ N}$

To determine

(c)

To Calculate:

The magnitude of the force on the alpha particle when it is moving in the positive z-direction in a magnetic field parallel to the positive x-axis.

Answer to Problem 11.1CYU

The magnetic force on the alpha particle is $2.4\times {10}^{-14}$ N towards the positive y-axis.

Explanation of Solution

Given:

Charge on the alpha particle, $q=3.2\times {10}^{-19}C$

Magnetic field parallel to the positive x-axis, $B=1.5\text{ T}$

The velocity of the alpha particle in the positive z-direction, $\overrightarrow{v}=5\times {10}^4\text{ <mover accent="true"><mo>k</mo><mo>⌢</mo></mover><mo> m/s</mo>}$

Formula used:

The magnetic force on a charged particle is given as

  ${\overrightarrow{F}}_B=q(\overrightarrow{v}\times \overrightarrow{B})$ or $F_B=qvB\sin \theta$ .-----(1)

Here $q$ is the charge of the particle, $v$ is the velocity of the particle, $B$ is the magnetic field

& $\theta$ is the angle between the velocity vector and the magnetic field vector.

Calculation:

The magnetic force on the alpha particle is given by

  $\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$

Substituting the value of $q$, $\overrightarrow{v}$ & $\overrightarrow{B}$ we get,

  $\overrightarrow{F}=(3.2\times {10}^{-19})\left[(5\times {10}^4\text{ <mover accent="true"><mo>k</mo><mo>⌢</mo></mover>})\times (1.5\stackrel{⌢}{i})\right]$

Or $\overrightarrow{F}=2.4\times {10}^{-14}\text{<mover accent="true"><mo>j</mo><mo>⌢</mo></mover><mo> N</mo>}$

To determine

(d)

To Calculate:

The magnitude of the force on the alpha particle when it is moving in an arbitrary direction in a magnetic field parallel to the positive x-axis.

Answer to Problem 11.1CYU

The magnetic force on the alpha particle is $1.52\times {10}^{-7}\text{ N}$.

Explanation of Solution

Given:

Charge on the alpha particle, $q=3.2\times {10}^{-19}C$

Magnetic field parallel to the positive x-axis, $B=1.5\text{ T}$

Velocity of the alpha particle, $\overrightarrow{v}=(2\stackrel{⌢}{i}-3\stackrel{⌢}{j}+1\stackrel{⌢}{k})\times {10}^4\text{ m/s}$

Formula used:

The magnetic force on a charged particle is given as

  ${\overrightarrow{F}}_B=q(\overrightarrow{v}\times \overrightarrow{B})$ or $F_B=qvB\sin \theta$ .-----(1)

Here $q$ is the charge of the particle, $v$ is the velocity of the particle, $B$ is the magnetic field

& $\theta$ is the angle between the velocity vector and the magnetic field vector.

Calculation:

The magnetic force on the alpha particle is given by

  $\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$

Substituting the value of $q$, $\overrightarrow{v}$ & $\overrightarrow{B}$ we get,

  $\begin{array}{l}\overrightarrow{F}=(3.2\times {10}^{-19})\left[((2\stackrel{⌢}{i}-3\stackrel{⌢}{j}+1\stackrel{⌢}{k})\times {10}^4)\times (1.5\stackrel{⌢}{i})\right]\\ \end{array}$

Or $\overrightarrow{F}=(1.44\stackrel{⌢}{k}+0.48\stackrel{⌢}{j})\times {10}^{-14}\text{ N}$

The magnitude of the force is $\left|\overrightarrow{F}\right|=\sqrt{\left({(1.44)}^2+{(0.48)}^2\right)\times {10}^{-14}}=1.52\times {10}^{-7}\text{ N}$