General Physics, 2nd Edition
General Physics, 2nd Edition
2nd Edition
ISBN: 9780471522782
Author: Morton M. Sternheim
Publisher: WILEY
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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
Check Mark

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
Check Mark

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
Check Mark

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
Check Mark

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
Check Mark

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
Check Mark

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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