Chapter 19, Problem 5E

(a)

To determine

The magnitude and direction of acceleration of a charged particle in Figure 19.33 if $v$ is in the $+x\text{ direction}$.

Answer to Problem 5E

Both magnitude and direction of acceleration is zero when $v$ is in the $\underset{\_}{+x\text{ direction}}$.

Explanation of Solution

Moving electric charges experience forces near a magnetic field while a charge at rest experience no such forces.

Write the expression to find the magnetic force experiencing on a moving charge.

    $F=\left|q\right|vB\sin \theta$        (I)

Here, $F$ is the magnetic force, $q$ is the charge of the particle, $v$ is the velocity of the charge, $B$ is the magnitude of the magnetic field, and $\theta$ is the angle between magnetic field and velocity of the charge.

From Newton’s second law of motion,

    $F=ma$        (II)

Here, $m$ is the mass of the particle, and $a$ is the acceleration.

From equation (II) magnitude of acceleration of the particle can be deducted as,

    $a=\frac{F}{m}$        (III)

Substitute equation (III) in (I) to find the acceleration of the particle.

    $a=\left(\frac{q}{m}\right)vB\sin \theta$        (IV)

Conclusion:

If $v$ is in the $+x\text{ direction}$ then the velocity and magnetic field will be parallel to each other.

Substitute $\theta =0°$ in equation (IV) to find acceleration.

    $\begin{array}{c}a=\left(\frac{q}{m}\right)vB\sin 0°\\ =0\end{array}$

Therefore, both magnitude and direction of acceleration is zero when $v$ is in the $\underset{\_}{+x\text{ direction}}$.

(b)

To determine

The magnitude and direction of acceleration of a charged particle in Figure 19.33 if $v$ is in the $-x\text{ direction}$.

Answer to Problem 5E

Both magnitude and direction of acceleration is zero when $v$ is in the $\underset{\_}{-x\text{ direction}}$.

Explanation of Solution

Equation (IV) in part (a) provides the magnitude and direction of acceleration of the particle. If $v$ is in the $-x\text{ direction}$ then the velocity and magnetic field will be anti-parallel to each other.

Conclusion:

Substitute $\theta =180°$ in equation (IV) to find acceleration.

    $\begin{array}{c}a=\left(\frac{q}{m}\right)vB\sin 180°\\ =0\end{array}$

Therefore, both magnitude and direction of acceleration is zero when $v$ is in the $\underset{\_}{-x\text{ direction}}$.

(c)

To determine

The magnitude and direction of acceleration of a charged particle in Figure 19.33 if $v$ is in the $+y\text{ direction}$.

Answer to Problem 5E

The magnitude of the force is $\underset{\_}{\left(qvB\right)/m}$, and direction will point into the page since the charge is moving in $+y\text{ direction}$.

Explanation of Solution

Equation (IV) in part a provides the magnitude and direction of acceleration of the particle. If $v$ is in the $+y\text{ direction}$ then the velocity and magnetic field will be perpendicular to each other.

Conclusion:

Substitute $\theta =90°$ in equation (IV) to find acceleration.

    $\begin{array}{c}a=\left(\frac{q}{m}\right)vB\sin 90°\\ =\frac{\left|qvB\right|}{m}\end{array}$

Use right hand rule to find direction of acceleration. Direction of acceleration will be same as that of the force. Since $B$ is in $+x\text{ direction}$ and $v$ is in $+y\text{ direction}$, acceleration will point into the page.

Therefore, magnitude of the force is $\underset{\_}{\left(qvB\right)/m}$, and direction will point into the page since the charge is moving in $+y\text{ direction}$.

(d)

To determine

The magnitude and direction of acceleration of a charged particle in Figure 19.33 if $v$ is in the $-y\text{ direction}$.

Answer to Problem 5E

The magnitude of the force is $\underset{\_}{\left(qvB\right)/m}$, and direction will point out of the page since the charge is moving in $-y\text{ direction}$.

Explanation of Solution

Equation (IV) in part (a) provides the magnitude and direction of acceleration of the particle. If $v$ is in the $-y\text{ direction}$ then the velocity and magnetic field will be perpendicular to each other.

Conclusion:

Substitute $\theta =-90°$ in equation (IV) to find acceleration.

    $\begin{array}{c}a=\left(\frac{q}{m}\right)vB\sin \left(-90°\right)\\ =-\frac{\left|qvB\right|}{m}\end{array}$

Here, $-\text{ve}$ sign implies direction.

Use right hand rule to find direction of acceleration. Direction of acceleration will be same as that of the force. Since $B$ is in $+x\text{ direction}$ and $v$ is in $-y\text{ direction}$, acceleration will point out of the page.

Therefore, the magnitude the force is $\underset{\_}{\left(qvB\right)/m}$, and direction will point out of the page since the charge is moving in $-y\text{ direction}$.

(e)

To determine

The magnitude and direction of acceleration of a charged particle in Figure 19.33 if $v$ is out of the page.

Answer to Problem 5E

The magnitude of the force is $\underset{\_}{\left(qvB\right)/m}$, and direction will be in the $\underset{\_}{+y\text{ direction}}$ when $v$ is out of the page.

Explanation of Solution

Equation (IV) in part (a) provides the magnitude and direction of acceleration of the particle. If $v$ is out of the page then the velocity and magnetic field will be perpendicular to each other.

Conclusion:

Substitute $\theta =90°$ in equation (IV) to find acceleration.

    $\begin{array}{c}a=\left(\frac{q}{m}\right)vB\sin \left(90°\right)\\ =\frac{\left|qvB\right|}{m}\end{array}$

Use right hand rule to find direction of acceleration. Direction of acceleration will be same as that of force. Since $B$ is in $+x\text{ direction}$ and $v$ is out of the page, acceleration will be in $+y\text{ direction}$.

Therefore, the magnitude of the force is $\underset{\_}{\left(qvB\right)/m}$, and direction will be in $\underset{\_}{+y\text{ direction}}$.

(f)

To determine

The magnitude and direction of acceleration of a charged particle in Figure 19.33 if $v$ is into the page.

Answer to Problem 5E

The magnitude of the force is $\underset{\_}{\left(qvB\right)/m}$, and direction will be in the $\underset{\_}{-y\text{ direction}}$ when $v$ is into of the page.

Explanation of Solution

Equation (IV) in part (a) provides the magnitude and direction of acceleration of the particle. If $v$ is into of the page direction then the velocity and magnetic field will be perpendicular to each other.

Conclusion:

Substitute $\theta =-90°$ in equation (IV) to find acceleration.

    $\begin{array}{c}a=\left(\frac{q}{m}\right)vB\sin \left(-90°\right)\\ =-\frac{\left|qvB\right|}{m}\end{array}$

Use right hand rule to find direction of acceleration. Direction of acceleration will be same as that of force. Since $B$ is in $+x\text{ direction}$ and $v$ is into of the page, acceleration will be in $-y\text{ direction}$.

Therefore, the magnitude of the force is $\underset{\_}{\left(qvB\right)/m}$, and direction will be in $\underset{\_}{-y\text{ direction}}$.