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Chapter 27: Problem 3

In a 1.25 -T magnetic field directed vertically upward, a particle having a charge of magnitude 8.50\(\mu \mathrm{C}\) and initially moving northward at 4.75 \(\mathrm{km} / \mathrm{s}\) is deflected toward the east. (a) What is the sign of the charge of this particle? Make a sketch to illustrate how you found your answer. (b) Find the magnetic force on the particle.

Short Answer

Expert verified
(a) The particle is positively charged. (b) The magnetic force is 50.5 N east.

Step by step solution

01

Determine the Direction of Force

According to the right-hand rule, when a positive charge moves through a magnetic field, the force exerted on it is perpendicular to both the direction of the magnetic field and the velocity of the charge. Since the particle is deflected toward the east when moving northward in an upward magnetic field, the force is to the right (to the east), indicating the particle has a positive charge.
02

Identify Sign of the Charge

Since the right-hand rule prediction matched the actual deflection, the charge must be positive. If the charge were negative, it would deflect in the opposite direction (west), indicating that the sign of the charge is positive.
03

Calculate Magnetic Force

The magnetic force on a charged particle moving through a magnetic field is given by the formula: \( F = qvB \sin(\theta) \), where \( q \) is the charge, \( v \) is the velocity, \( B \) is the magnetic field strength, and \( \theta \) is the angle between \( v \) and \( B \). Here, \( \theta = 90^{\circ} \), so \( \sin(\theta) = 1 \). Substitute the given values: \( q = 8.50 \times 10^{-6} \; \mathrm{C} \), \( v = 4.75 \times 10^{3} \; \mathrm{m/s} \), and \( B = 1.25 \; \mathrm{T} \):\[F = (8.50 \times 10^{-6} \; \mathrm{C})(4.75 \times 10^{3} \; \mathrm{m/s})(1.25 \; \mathrm{T}) (1) = 50.5 \; \mathrm{N}.\]
04

Present the Results

(a) The particle has a positive charge. (b) The magnetic force on the particle is 50.5 N directed east.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Right-Hand Rule
The Right-Hand Rule is a handy tool used in physics to determine the direction of force exerted by a magnetic field on a moving charge. Imagine bending the fingers of your right hand around so that your thumb, index finger, and middle finger are perpendicular to each other, like the "axes" on a 3D graph.

- Your thumb represents the direction of the velocity of a positive charge.
- Your index finger points in the direction of the magnetic field lines.
- Your middle finger, when extended, then indicates the direction of the force experienced by a positive charge.

In this problem, the charge is moving north, so you point your thumb in the north direction. The magnetic field is directed upwards, so your index finger points upwards. When you do this, your middle finger will naturally point to the right, which means east. Hence, a positive charge deflects to the east. This indicates that since the deflected direction (east) matches the Right-Hand Rule for a positive charge, the charge of the particle is indeed positive.
Charge Sign Determination
Charge Sign Determination involves understanding the deflection direction of a charged particle in a magnetic field. If a charged particle moving through a magnetic field gets deflected, the direction of this deflection helps determine the sign of the charge. Given that the particle deflects east from a northward initial path, and the magnetic field is upwards, we use the Right-Hand Rule again:

- Align your thumb with the particle's velocity (northward).
- Point your index finger in the direction of the magnetic field (upward).
- See which way your middle finger points (east), if you're using your right hand correctly.

This deflection direction matching the Right-Hand Rule implies a positive charge. If the charge were negative, the force direction would be opposite, deflecting west instead. Thus, the charge on the particle is positive.
Magnetic Field Influence
Magnetic Field Influence refers to the impact that a magnetic field has on a moving charged particle. The extent of this influence is calculated using the formula for magnetic force: \[ F = qvB \sin(\theta) \]where- \( F \) is the magnetic force,
- \( q \) is the charge of the particle,
- \( v \) is the velocity,
- \( B \) is the magnetic field strength, and
- \( \theta \) is the angle between the velocity and the magnetic field direction.In this scenario, the angle \( \theta \) is \( 90^{\circ} \) because the velocity and magnetic field are perpendicular, making \( \sin(\theta) = 1 \). The force is then determined entirely by multiplying \( q, v, \) and \( B \):- \( q = 8.50 \times 10^{-6} \; \mathrm{C} \)
- \( v = 4.75 \times 10^{3} \; \mathrm{m/s} \)
- \( B = 1.25 \; \mathrm{T} \)Substituting these values yields:\[ F = (8.50 \times 10^{-6} \; \mathrm{C})(4.75 \times 10^{3} \; \mathrm{m/s})(1.25 \; \mathrm{T}) = 50.5 \; \mathrm{N} \]The particle hence experiences a magnetic force of 50.5 N, directing it toward the east.

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Most popular questions from this chapter

A particle with a charge of \(-1.24 \times 10^{-8} \mathrm{C}\) is moving with instantaneous velocity \(\vec{v}=\left(4.19 \times 10^{4} \mathrm{m} / \mathrm{s}\right) \hat{\imath}+(-3.85 \times\) \(10^{4} \mathrm{m} / \mathrm{s} ) \hat{\boldsymbol{J}}\) . What is the force exerted on this particle by a mag- netic field (a) \(\vec{\boldsymbol{B}}=(1.40 \mathrm{T}) \hat{\boldsymbol{i}}\) and \((\mathrm{b}) \vec{\boldsymbol{B}}=(1.40 \mathrm{T}) \hat{\boldsymbol{k}} ?\)

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