Concept explainers
(a)
The magnitude and direction of acceleration of a charged particle in Figure 19.33 if
(a)
Answer to Problem 5E
Both magnitude and direction of acceleration is zero when
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.
Here,
From Newton’s second law of motion,
Here,
From equation (II) magnitude of acceleration of the particle can be deducted as,
Substitute equation (III) in (I) to find the acceleration of the particle.
Conclusion:
If
Substitute
Therefore, both magnitude and direction of acceleration is zero when
(b)
The magnitude and direction of acceleration of a charged particle in Figure 19.33 if
(b)
Answer to Problem 5E
Both magnitude and direction of acceleration is zero when
Explanation of Solution
Equation (IV) in part (a) provides the magnitude and direction of acceleration of the particle. If
Conclusion:
Substitute
Therefore, both magnitude and direction of acceleration is zero when
(c)
The magnitude and direction of acceleration of a charged particle in Figure 19.33 if
(c)
Answer to Problem 5E
The magnitude of the force is
Explanation of Solution
Equation (IV) in part a provides the magnitude and direction of acceleration of the particle. If
Conclusion:
Substitute
Use right hand rule to find direction of acceleration. Direction of acceleration will be same as that of the force. Since
Therefore, magnitude of the force is
(d)
The magnitude and direction of acceleration of a charged particle in Figure 19.33 if
(d)
Answer to Problem 5E
The magnitude of the force is
Explanation of Solution
Equation (IV) in part (a) provides the magnitude and direction of acceleration of the particle. If
Conclusion:
Substitute
Here,
Use right hand rule to find direction of acceleration. Direction of acceleration will be same as that of the force. Since
Therefore, the magnitude the force is
(e)
The magnitude and direction of acceleration of a charged particle in Figure 19.33 if
(e)
Answer to Problem 5E
The magnitude of the force is
Explanation of Solution
Equation (IV) in part (a) provides the magnitude and direction of acceleration of the particle. If
Conclusion:
Substitute
Use right hand rule to find direction of acceleration. Direction of acceleration will be same as that of force. Since
Therefore, the magnitude of the force is
(f)
The magnitude and direction of acceleration of a charged particle in Figure 19.33 if
(f)
Answer to Problem 5E
The magnitude of the force is
Explanation of Solution
Equation (IV) in part (a) provides the magnitude and direction of acceleration of the particle. If
Conclusion:
Substitute
Use right hand rule to find direction of acceleration. Direction of acceleration will be same as that of force. Since
Therefore, the magnitude of the force is
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Chapter 19 Solutions
General Physics, 2nd Edition
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- 12.4 Check Your Understanding Two wires, both carrying current out of the page, have a current of magnitude 2.0 mA and 3.0 mA, respectively. The first wire is located at (0.0 cm, 5.0 cm) while the other wire is located at (12.0 cm, 0.0 cm). What is the magnitude of the magnetic force per unit length of the first wire on the second and the second wire on the first?arrow_forward, A proton, deuteron, and an alpha-particle ae all accelerated from rest through the same potential difference. They then enter the same magnetic field, moving perpendicular to it. Compute the ratios of the radii of their circular paths. Assume that md= 2wmp and ma= 4mp.arrow_forwardWhen the current through a circular loop is 6.0 A, the magnetic field at its center is 2.0104 T. What is the radius of the loop?arrow_forward
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