The magnitude and direction of acceleration of a charged particle in Figure 19.33 if v is in the + x direction .

Question

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Answer 1

Question

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 $+x direction$ .

(a)

Expert Solution

Answer to Problem 5E

Both magnitude and direction of acceleration is zero when $v$ is in the $+x direction_$ .

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.

$F=|q|vBsinθ$ (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 $θ$ 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,

$a=Fm$ (III)

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

$a=(qm)vBsinθ$ (IV)

Conclusion:

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

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

$a=(qm)vBsin0°=0$

Therefore, both magnitude and direction of acceleration is zero when $v$ is in the $+x direction_$ .

(b)

To determine

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

(b)

Expert Solution

Answer to Problem 5E

Both magnitude and direction of acceleration is zero when $v$ is in the $−x direction_$ .

Explanation of Solution

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

Conclusion:

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

$a=(qm)vBsin180°=0$

Therefore, both magnitude and direction of acceleration is zero when $v$ is in the $−x direction_$ .

(c)

To determine

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

(c)

Expert Solution

Answer to Problem 5E

The magnitude of the force is $(qvB)/m_$ , and direction will point into the page since the charge is moving in $+y direction$ .

Explanation of Solution

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

Conclusion:

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

$a=(qm)vBsin90°=|qvB|m$

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

Therefore, magnitude of the force is $(qvB)/m_$ , and direction will point into the page since the charge is moving in $+y direction$ .

(d)

To determine

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

(d)

Expert Solution

Answer to Problem 5E

The magnitude of the force is $(qvB)/m_$ , and direction will point out of the page since the charge is moving in $−y direction$ .

Explanation of Solution

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

Conclusion:

Substitute $θ=−90°$ in equation (IV) to find acceleration.

$a=(qm)vBsin(−90°)=−|qvB|m$

Here, $−ve$ 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 $+x direction$ and $v$ is in $−y direction$ , acceleration will point out of the page.

Therefore, the magnitude the force is $(qvB)/m_$ , and direction will point out of the page since the charge is moving in $−y direction$ .

(e)

To determine

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

(e)

Expert Solution

Answer to Problem 5E

The magnitude of the force is $(qvB)/m_$ , and direction will be in the $+y direction_$ 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 $θ=90°$ in equation (IV) to find acceleration.

$a=(qm)vBsin(90°)=|qvB|m$

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

Therefore, the magnitude of the force is $(qvB)/m_$ , and direction will be in $+y direction_$ .

(f)

To determine

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

(f)

Expert Solution

Answer to Problem 5E

The magnitude of the force is $(qvB)/m_$ , and direction will be in the $−y direction_$ 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 $θ=−90°$ in equation (IV) to find acceleration.

$a=(qm)vBsin(−90°)=−|qvB|m$

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

Therefore, the magnitude of the force is $(qvB)/m_$ , and direction will be in $−y direction_$ .

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Answer 2

Question

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 $+x direction$ .

(a)

Expert Solution

Answer to Problem 5E

Both magnitude and direction of acceleration is zero when $v$ is in the $+x direction_$ .

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.

$F=|q|vBsinθ$ (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 $θ$ 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,

$a=Fm$ (III)

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

$a=(qm)vBsinθ$ (IV)

Conclusion:

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

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

$a=(qm)vBsin0°=0$

Therefore, both magnitude and direction of acceleration is zero when $v$ is in the $+x direction_$ .

(b)

To determine

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

(b)

Expert Solution

Answer to Problem 5E

Both magnitude and direction of acceleration is zero when $v$ is in the $−x direction_$ .

Explanation of Solution

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

Conclusion:

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

$a=(qm)vBsin180°=0$

Therefore, both magnitude and direction of acceleration is zero when $v$ is in the $−x direction_$ .

(c)

To determine

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

(c)

Expert Solution

Answer to Problem 5E

The magnitude of the force is $(qvB)/m_$ , and direction will point into the page since the charge is moving in $+y direction$ .

Explanation of Solution

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

Conclusion:

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

$a=(qm)vBsin90°=|qvB|m$

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

Therefore, magnitude of the force is $(qvB)/m_$ , and direction will point into the page since the charge is moving in $+y direction$ .

(d)

To determine

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

(d)

Expert Solution

Answer to Problem 5E

The magnitude of the force is $(qvB)/m_$ , and direction will point out of the page since the charge is moving in $−y direction$ .

Explanation of Solution

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

Conclusion:

Substitute $θ=−90°$ in equation (IV) to find acceleration.

$a=(qm)vBsin(−90°)=−|qvB|m$

Here, $−ve$ 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 $+x direction$ and $v$ is in $−y direction$ , acceleration will point out of the page.

Therefore, the magnitude the force is $(qvB)/m_$ , and direction will point out of the page since the charge is moving in $−y direction$ .

(e)

To determine

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

(e)

Expert Solution

Answer to Problem 5E

The magnitude of the force is $(qvB)/m_$ , and direction will be in the $+y direction_$ 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 $θ=90°$ in equation (IV) to find acceleration.

$a=(qm)vBsin(90°)=|qvB|m$

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

Therefore, the magnitude of the force is $(qvB)/m_$ , and direction will be in $+y direction_$ .

(f)

To determine

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

(f)

Expert Solution

Answer to Problem 5E

The magnitude of the force is $(qvB)/m_$ , and direction will be in the $−y direction_$ 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 $θ=−90°$ in equation (IV) to find acceleration.

$a=(qm)vBsin(−90°)=−|qvB|m$

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

Therefore, the magnitude of the force is $(qvB)/m_$ , and direction will be in $−y direction_$ .

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Answer 3

Question

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 $+x direction$ .

(a)

Expert Solution

Answer to Problem 5E

Both magnitude and direction of acceleration is zero when $v$ is in the $+x direction_$ .

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.

$F=|q|vBsinθ$ (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 $θ$ 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,

$a=Fm$ (III)

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

$a=(qm)vBsinθ$ (IV)

Conclusion:

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

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

$a=(qm)vBsin0°=0$

Therefore, both magnitude and direction of acceleration is zero when $v$ is in the $+x direction_$ .

(b)

To determine

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

(b)

Expert Solution

Answer to Problem 5E

Both magnitude and direction of acceleration is zero when $v$ is in the $−x direction_$ .

Explanation of Solution

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

Conclusion:

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

$a=(qm)vBsin180°=0$

Therefore, both magnitude and direction of acceleration is zero when $v$ is in the $−x direction_$ .

(c)

To determine

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

(c)

Expert Solution

Answer to Problem 5E

The magnitude of the force is $(qvB)/m_$ , and direction will point into the page since the charge is moving in $+y direction$ .

Explanation of Solution

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

Conclusion:

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

$a=(qm)vBsin90°=|qvB|m$

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

Therefore, magnitude of the force is $(qvB)/m_$ , and direction will point into the page since the charge is moving in $+y direction$ .

(d)

To determine

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

(d)

Expert Solution

Answer to Problem 5E

The magnitude of the force is $(qvB)/m_$ , and direction will point out of the page since the charge is moving in $−y direction$ .

Explanation of Solution

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

Conclusion:

Substitute $θ=−90°$ in equation (IV) to find acceleration.

$a=(qm)vBsin(−90°)=−|qvB|m$

Here, $−ve$ 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 $+x direction$ and $v$ is in $−y direction$ , acceleration will point out of the page.

Therefore, the magnitude the force is $(qvB)/m_$ , and direction will point out of the page since the charge is moving in $−y direction$ .

(e)

To determine

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

(e)

Expert Solution

Answer to Problem 5E

The magnitude of the force is $(qvB)/m_$ , and direction will be in the $+y direction_$ 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 $θ=90°$ in equation (IV) to find acceleration.

$a=(qm)vBsin(90°)=|qvB|m$

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

Therefore, the magnitude of the force is $(qvB)/m_$ , and direction will be in $+y direction_$ .

(f)

To determine

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

(f)

Expert Solution

Answer to Problem 5E

The magnitude of the force is $(qvB)/m_$ , and direction will be in the $−y direction_$ 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 $θ=−90°$ in equation (IV) to find acceleration.

$a=(qm)vBsin(−90°)=−|qvB|m$

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

Therefore, the magnitude of the force is $(qvB)/m_$ , and direction will be in $−y direction_$ .

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Answer 4

Question

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 $+x direction$ .

(a)

Expert Solution

Answer to Problem 5E

Both magnitude and direction of acceleration is zero when $v$ is in the $+x direction_$ .

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.

$F=|q|vBsinθ$ (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 $θ$ 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,

$a=Fm$ (III)

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

$a=(qm)vBsinθ$ (IV)

Conclusion:

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

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

$a=(qm)vBsin0°=0$

Therefore, both magnitude and direction of acceleration is zero when $v$ is in the $+x direction_$ .

(b)

To determine

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

(b)

Expert Solution

Answer to Problem 5E

Both magnitude and direction of acceleration is zero when $v$ is in the $−x direction_$ .

Explanation of Solution

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

Conclusion:

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

$a=(qm)vBsin180°=0$

Therefore, both magnitude and direction of acceleration is zero when $v$ is in the $−x direction_$ .

(c)

To determine

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

(c)

Expert Solution

Answer to Problem 5E

The magnitude of the force is $(qvB)/m_$ , and direction will point into the page since the charge is moving in $+y direction$ .

Explanation of Solution

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

Conclusion:

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

$a=(qm)vBsin90°=|qvB|m$

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

Therefore, magnitude of the force is $(qvB)/m_$ , and direction will point into the page since the charge is moving in $+y direction$ .

(d)

To determine

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

(d)

Expert Solution

Answer to Problem 5E

The magnitude of the force is $(qvB)/m_$ , and direction will point out of the page since the charge is moving in $−y direction$ .

Explanation of Solution

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

Conclusion:

Substitute $θ=−90°$ in equation (IV) to find acceleration.

$a=(qm)vBsin(−90°)=−|qvB|m$

Here, $−ve$ 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 $+x direction$ and $v$ is in $−y direction$ , acceleration will point out of the page.

Therefore, the magnitude the force is $(qvB)/m_$ , and direction will point out of the page since the charge is moving in $−y direction$ .

(e)

To determine

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

(e)

Expert Solution

Answer to Problem 5E

The magnitude of the force is $(qvB)/m_$ , and direction will be in the $+y direction_$ 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 $θ=90°$ in equation (IV) to find acceleration.

$a=(qm)vBsin(90°)=|qvB|m$

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

Therefore, the magnitude of the force is $(qvB)/m_$ , and direction will be in $+y direction_$ .

(f)

To determine

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

(f)

Expert Solution

Answer to Problem 5E

The magnitude of the force is $(qvB)/m_$ , and direction will be in the $−y direction_$ 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 $θ=−90°$ in equation (IV) to find acceleration.

$a=(qm)vBsin(−90°)=−|qvB|m$

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

Therefore, the magnitude of the force is $(qvB)/m_$ , and direction will be in $−y direction_$ .

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Answer 5

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 $+x direction$ .

(a)

Expert Solution

Answer to Problem 5E

Both magnitude and direction of acceleration is zero when $v$ is in the $+x direction_$ .

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.

$F=|q|vBsinθ$ (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 $θ$ 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,

$a=Fm$ (III)

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

$a=(qm)vBsinθ$ (IV)

Conclusion:

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

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

$a=(qm)vBsin0°=0$

Therefore, both magnitude and direction of acceleration is zero when $v$ is in the $+x direction_$ .

(b)

To determine

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

(b)

Expert Solution

Answer to Problem 5E

Both magnitude and direction of acceleration is zero when $v$ is in the $−x direction_$ .

Explanation of Solution

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

Conclusion:

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

$a=(qm)vBsin180°=0$

Therefore, both magnitude and direction of acceleration is zero when $v$ is in the $−x direction_$ .

(c)

To determine

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

(c)

Expert Solution

Answer to Problem 5E

The magnitude of the force is $(qvB)/m_$ , and direction will point into the page since the charge is moving in $+y direction$ .

Explanation of Solution

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

Conclusion:

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

$a=(qm)vBsin90°=|qvB|m$

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

Therefore, magnitude of the force is $(qvB)/m_$ , and direction will point into the page since the charge is moving in $+y direction$ .

(d)

To determine

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

(d)

Expert Solution

Answer to Problem 5E

The magnitude of the force is $(qvB)/m_$ , and direction will point out of the page since the charge is moving in $−y direction$ .

Explanation of Solution

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

Conclusion:

Substitute $θ=−90°$ in equation (IV) to find acceleration.

$a=(qm)vBsin(−90°)=−|qvB|m$

Here, $−ve$ 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 $+x direction$ and $v$ is in $−y direction$ , acceleration will point out of the page.

Therefore, the magnitude the force is $(qvB)/m_$ , and direction will point out of the page since the charge is moving in $−y direction$ .

(e)

To determine

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

(e)

Expert Solution

Answer to Problem 5E

The magnitude of the force is $(qvB)/m_$ , and direction will be in the $+y direction_$ 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 $θ=90°$ in equation (IV) to find acceleration.

$a=(qm)vBsin(90°)=|qvB|m$

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

Therefore, the magnitude of the force is $(qvB)/m_$ , and direction will be in $+y direction_$ .

(f)

To determine

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

(f)

Expert Solution

Answer to Problem 5E

The magnitude of the force is $(qvB)/m_$ , and direction will be in the $−y direction_$ 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 $θ=−90°$ in equation (IV) to find acceleration.

$a=(qm)vBsin(−90°)=−|qvB|m$

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

Therefore, the magnitude of the force is $(qvB)/m_$ , and direction will be in $−y direction_$ .

Answer 6

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 $+x direction$ .

(a)

Expert Solution

Answer to Problem 5E

Both magnitude and direction of acceleration is zero when $v$ is in the $+x direction_$ .

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.

$F=|q|vBsinθ$ (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 $θ$ 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,

$a=Fm$ (III)

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

$a=(qm)vBsinθ$ (IV)

Conclusion:

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

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

$a=(qm)vBsin0°=0$

Therefore, both magnitude and direction of acceleration is zero when $v$ is in the $+x direction_$ .

Answer 7

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 $+x direction$ .

Answer 8

(a)

Expert Solution

Answer to Problem 5E

Both magnitude and direction of acceleration is zero when $v$ is in the $+x direction_$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

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.

$F=|q|vBsinθ$ (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 $θ$ 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,

$a=Fm$ (III)

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

$a=(qm)vBsinθ$ (IV)

Conclusion:

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

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

$a=(qm)vBsin0°=0$

Therefore, both magnitude and direction of acceleration is zero when $v$ is in the $+x direction_$ .

Answer 13

(b)

To determine

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

(b)

Expert Solution

Answer to Problem 5E

Both magnitude and direction of acceleration is zero when $v$ is in the $−x direction_$ .

Explanation of Solution

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

Conclusion:

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

$a=(qm)vBsin180°=0$

Therefore, both magnitude and direction of acceleration is zero when $v$ is in the $−x direction_$ .

Answer 14

(b)

To determine

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

Answer 15

(b)

Expert Solution

Answer to Problem 5E

Both magnitude and direction of acceleration is zero when $v$ is in the $−x direction_$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

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

Conclusion:

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

$a=(qm)vBsin180°=0$

Therefore, both magnitude and direction of acceleration is zero when $v$ is in the $−x direction_$ .

Answer 20

(c)

To determine

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

(c)

Expert Solution

Answer to Problem 5E

The magnitude of the force is $(qvB)/m_$ , and direction will point into the page since the charge is moving in $+y direction$ .

Explanation of Solution

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

Conclusion:

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

$a=(qm)vBsin90°=|qvB|m$

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

Therefore, magnitude of the force is $(qvB)/m_$ , and direction will point into the page since the charge is moving in $+y direction$ .

Answer 21

(c)

To determine

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

Answer 22

(c)

Expert Solution

Answer to Problem 5E

The magnitude of the force is $(qvB)/m_$ , and direction will point into the page since the charge is moving in $+y direction$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

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

Conclusion:

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

$a=(qm)vBsin90°=|qvB|m$

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

Therefore, magnitude of the force is $(qvB)/m_$ , and direction will point into the page since the charge is moving in $+y direction$ .

Answer 27

(d)

To determine

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

(d)

Expert Solution

Answer to Problem 5E

The magnitude of the force is $(qvB)/m_$ , and direction will point out of the page since the charge is moving in $−y direction$ .

Explanation of Solution

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

Conclusion:

Substitute $θ=−90°$ in equation (IV) to find acceleration.

$a=(qm)vBsin(−90°)=−|qvB|m$

Here, $−ve$ 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 $+x direction$ and $v$ is in $−y direction$ , acceleration will point out of the page.

Therefore, the magnitude the force is $(qvB)/m_$ , and direction will point out of the page since the charge is moving in $−y direction$ .

Answer 28

(d)

To determine

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

Answer 29

(d)

Expert Solution

Answer to Problem 5E

The magnitude of the force is $(qvB)/m_$ , and direction will point out of the page since the charge is moving in $−y direction$ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

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

Conclusion:

Substitute $θ=−90°$ in equation (IV) to find acceleration.

$a=(qm)vBsin(−90°)=−|qvB|m$

Here, $−ve$ 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 $+x direction$ and $v$ is in $−y direction$ , acceleration will point out of the page.

Therefore, the magnitude the force is $(qvB)/m_$ , and direction will point out of the page since the charge is moving in $−y direction$ .

Answer 34

(e)

To determine

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

(e)

Expert Solution

Answer to Problem 5E

The magnitude of the force is $(qvB)/m_$ , and direction will be in the $+y direction_$ 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 $θ=90°$ in equation (IV) to find acceleration.

$a=(qm)vBsin(90°)=|qvB|m$

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

Therefore, the magnitude of the force is $(qvB)/m_$ , and direction will be in $+y direction_$ .

Answer 35

(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 36

(e)

Expert Solution

Answer to Problem 5E

The magnitude of the force is $(qvB)/m_$ , and direction will be in the $+y direction_$ when $v$ is out of the page.

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

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 $θ=90°$ in equation (IV) to find acceleration.

$a=(qm)vBsin(90°)=|qvB|m$

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

Therefore, the magnitude of the force is $(qvB)/m_$ , and direction will be in $+y direction_$ .

Answer 41

(f)

To determine

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

(f)

Expert Solution

Answer to Problem 5E

The magnitude of the force is $(qvB)/m_$ , and direction will be in the $−y direction_$ 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 $θ=−90°$ in equation (IV) to find acceleration.

$a=(qm)vBsin(−90°)=−|qvB|m$

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

Therefore, the magnitude of the force is $(qvB)/m_$ , and direction will be in $−y direction_$ .

Answer 42

(f)

To determine

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

Answer 43

(f)

Expert Solution

Answer to Problem 5E

The magnitude of the force is $(qvB)/m_$ , and direction will be in the $−y direction_$ when $v$ is into of the page.

Answer 44

(f)

Expert Solution

Answer 45

(f)

Expert Solution

Answer 46

Expert Solution

Answer 47

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 $θ=−90°$ in equation (IV) to find acceleration.

$a=(qm)vBsin(−90°)=−|qvB|m$

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

Therefore, the magnitude of the force is $(qvB)/m_$ , and direction will be in $−y direction_$ .

Answer 48

Ch. 19 - Prob. 1RQ Ch. 19 - Prob. 2RQ Ch. 19 - Prob. 3RQ Ch. 19 - Prob. 4RQ Ch. 19 - Prob. 5RQ Ch. 19 - Prob. 6RQ Ch. 19 - Prob. 7RQ Ch. 19 - Prob. 8RQ Ch. 19 - Prob. 9RQ Ch. 19 - Prob. 10RQ

The magnitude and direction of acceleration of a charged particle in Figure 19.33 if v is in the + x direction .

Concept explainers

Videos

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

Answer to Problem 5E

Explanation of Solution

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

Answer to Problem 5E

Explanation of Solution

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

Answer to Problem 5E

Explanation of Solution

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

Answer to Problem 5E

Explanation of Solution

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

Explanation of Solution

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

Answer to Problem 5E

Explanation of Solution

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Chapter 19 Solutions

The magnitude and direction of acceleration of a charged particle in Figure 19.33 if v is in the + x direction .

Concept explainers

Videos

The magnitude and direction of acceleration of a charged particle in Figure 19.33 if v<m:math><m:mstyle mathvariant="bold" mathsize="normal"><m:mi>v</m:mi></m:mstyle></m:math> is in the +x direction<m:math><m:mrow><m:mo>+</m:mo><m:mi>x</m:mi><m:mtext> direction</m:mtext></m:mrow></m:math>.

Answer to Problem 5E

Explanation of Solution

The magnitude and direction of acceleration of a charged particle in Figure 19.33 if v<m:math><m:mstyle mathvariant="bold" mathsize="normal"><m:mi>v</m:mi></m:mstyle></m:math> is in the −x direction<m:math><m:mrow><m:mo>−</m:mo><m:mi>x</m:mi><m:mtext> direction</m:mtext></m:mrow></m:math>.

Answer to Problem 5E

Explanation of Solution

The magnitude and direction of acceleration of a charged particle in Figure 19.33 if v<m:math><m:mstyle mathvariant="bold" mathsize="normal"><m:mi>v</m:mi></m:mstyle></m:math> is in the +y direction<m:math><m:mrow><m:mo>+</m:mo><m:mi>y</m:mi><m:mtext> direction</m:mtext></m:mrow></m:math>.

Answer to Problem 5E

Explanation of Solution

The magnitude and direction of acceleration of a charged particle in Figure 19.33 if v<m:math><m:mstyle mathvariant="bold" mathsize="normal"><m:mi>v</m:mi></m:mstyle></m:math> is in the −y direction<m:math><m:mrow><m:mo>−</m:mo><m:mi>y</m:mi><m:mtext> direction</m:mtext></m:mrow></m:math>.

Answer to Problem 5E

Explanation of Solution

The magnitude and direction of acceleration of a charged particle in Figure 19.33 if v<m:math><m:mstyle mathvariant="bold" mathsize="normal"><m:mi>v</m:mi></m:mstyle></m:math> is out of the page.

Answer to Problem 5E

Explanation of Solution

The magnitude and direction of acceleration of a charged particle in Figure 19.33 if v<m:math><m:mstyle mathvariant="bold" mathsize="normal"><m:mi>v</m:mi></m:mstyle></m:math> is into the page.

Answer to Problem 5E

Explanation of Solution

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Chapter 19 Solutions

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

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

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

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

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

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