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Chapter 21: Problem 82

A particle has a charge of \(q=+5.60 \mu \mathrm{C}\) and is located at the coordinate origin. As the drawing shows, an electric field of \(E_{x}=+245 \mathrm{N} / \mathrm{C}\) exists along the \(+x\) axis. A magnetic field also exists, and its \(x\) and \(y\) components are \(B_{x}=+1.80 \mathrm{T}\) and \(B_{y}=+1.40 \mathrm{T} .\) Calculate the force (magnitude and direction) exerted on the particle by each of the three fields when it is (a) stationary, (b) moving along the \(+x\) axis at a speed of \(375 \mathrm{m} / \mathrm{s},\) and (c) moving along the \(+z\) axis at a speed of \(375 \mathrm{m} / \mathrm{s}\).

Short Answer

Expert verified
(a) 1.37 mN along +x. (b) 1.37 mN along +x, 2.94 mN in +z. (c) 1.37 mN along +x, 3.78 mN in -y direction.

Step by step solution

01

Identify forces for stationary particle

When the particle is stationary, the magnetic force is zero because the velocity is zero. Therefore, only the electric force acts on the particle. The electric force can be calculated using the formula: \[ F_e = q \times E \]Substitute the given values:\[ F_e = 5.60 \times 10^{-6} \times 245 \]Calculate to get \[ F_e = 1.372 \times 10^{-3} \text{ N} \].The direction of this force is along the +x axis since the electric field is in that direction.
02

Calculate forces when moving along +x axis

For a particle moving along the +x axis at speed 375 m/s, both electric and magnetic forces act on it. The electric force is the same as when stationary: \[ F_e = 1.372 \times 10^{-3} \text{ N} \].The magnetic force (if motion is along x) is calculated using:\[ F_m = qvB \sin \theta \]Where \( v = 375 \), \( B = 1.40 \) (only the y-component affects), and \( \theta = 90^\circ \):\[ F_m = 5.60 \times 10^{-6} \times 375 \times 1.40 = 2.94 \times 10^{-3} \text{ N} \].This force acts in the z direction due to the cross product nature \( \vec{v} \times \vec{B} \). The net magnetic force is perpendicular to the electric force in the z-direction.
03

Calculate forces when moving along +z axis

For a particle moving along the +z axis (375 m/s), the magnetic force results from both x and y components of the magnetic field.- Electric force remains as before, \( F_e = 1.372 \times 10^{-3} \text{ N} \).- For \( B_x = 1.80 \), use: \[ F_{m1} = qvB \sin \theta = 5.60 \times 10^{-6} \times 375 \times 1.80 = 3.78 \times 10^{-3} \text{ N} \] along -y.- Since \( B_y \) does not affect a z-axis velocity, \( F_{m2} \) for \( B_y \) is zero.Thus, the total magnetic force acts along the -y direction.

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Key Concepts

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

Electric Field
An electric field is a region where an electric charge experiences a force. This force is directly proportional to the magnitude of the electric field and the charge itself. The formula for calculating the electric force \( F_e \) acting on a charge \( q \) is \( F_e = q \times E \), where \( E \) is the electric field strength.
The direction of this force is determined by the direction of the electric field lines. In our exercise scenario, a positively charged particle experiences a force along the direction of the electric field. Since the electric field is oriented along the +x-axis, this simplifies the calculation because the force will also be parallel to the +x-axis.

In practical scenarios:
  • The presence of the electric field does not depend on a particle's motion.
  • The electric force acts immediately and is consistent as long as the field remains constant.
  • The electric force helps us understand phenomena like electrostatic interactions in atoms or the behavior in capacitors.
Magnetic Field
A magnetic field is a field produced by moving electric charges, which exerts a force on other moving charges. Note, the magnetic force depends on the velocity of the charge and both the magnitude and direction of the magnetic field. In our problem, the magnetic field has both x and y components given as \( B_x = 1.80 \) T and \( B_y = 1.40 \) T, respectively.
The magnetic force \( F_m \) is calculated using the cross product of velocity \( \vec{v} \) and the magnetic field \( \vec{B} \), given by the formula \( F_m = qvB \sin \theta \), where \( \theta \) is the angle between \( \vec{v} \) and \( \vec{B} \). This force is perpendicular to both the velocity and the magnetic field directions, which is why it can change the particle's trajectory without changing its speed.

Key points about magnetic fields:
  • A magnetic field alone cannot do work on a particle, as the force acts perpendicular to the velocity vector.
  • Magnetic forces are crucial in motors, generators, and magnetic storage devices.
  • In our scenario, different components of the particle's velocity result in forces in different directions. For example, moving along the x-axis affects the z-direction, while movement along the z-axis results in a force along the -y-axis due to different magnetic field interactions.
Particle Motion
The motion of a charged particle through electric and magnetic fields is key to understanding how forces act on it. When stationary, a charged particle only feels the electric field, which causes it to accelerate in the field's direction. For a positively charged particle, this means moving along the direction of the field lines.
With motion comes complexity. The magnetic force becomes significant only when the particle moves, as it relies on velocity. The direction and speed of this motion determine the nature of interaction with the magnetic field. For our problem:
  • When moving along the +x axis, the particle experiences both electric and magnetic forces. The magnetic component calculated is due to interaction with the y-component of the field, causing a perpendicular force in the z-direction.
  • When moving along the +z axis, the electric field still exerts a force in the +x direction, but the most interesting dynamical effect is the y-component magnetic force. This results in a force in the -y direction, as per our cross product calculation.
This way, particle motion in fields tells us how force orientations affect the path and requires careful calculation using established physical laws.

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Most popular questions from this chapter

The 1200 -turn coil in a dc motor has an area per turn of \(1.1 \times 10^{-2} \mathrm{m}^{2}\). The design for the motor specifies that the magnitude of the maximum torque is \(5.8 \mathrm{N} \cdot \mathrm{m}\) when the coil is placed in a \(0.20-\mathrm{T}\) magnetic field. What is the current in the coil?

In New England, the horizontal component of the earth's magnetic field has a magnitude of \(1.6 \times 10^{-5}\) T. An electron is shot vertically straight up from the ground with a speed of \(2.1 \times 10^{6} \mathrm{m} / \mathrm{s} .\) What is the magnitude of the acceleration caused by the magnetic force? Ignore the gravitational force acting on the electron.

A positively charged particle of mass \(7.2 \times 10^{-8} \mathrm{kg}\) is traveling due east with a speed of \(85 \mathrm{m} / \mathrm{s}\) and enters a \(0.31-\mathrm{T}\) uniform magnetic field. The particle moves through one- quarter of a circle in a time of \(2.2 \times 10^{-3} \mathrm{s},\) at which time it leaves the field heading due south. All during the motion the particle moves perpendicular to the magnetic field. (a) What is the magnitude of the magnetic force acting on the particle? (b) Determine the magnitude of its charge.

A long solenoid has 1400 turns per meter of length, and it carries a current of 3.5 A. A small circular coil of wire is placed inside the solenoid with the normal to the coil oriented at an angle of \(90.0^{\circ}\) with respect to the axis of the solenoid. The coil consists of 50 turns, has an area of \(1.2 \times 10^{-3} \mathrm{m}^{2}\) and carries a current of 0.50 A. Find the torque exerted on the coil.

The drawing shows a parallel plate capacitor that is moving with a speed of \(32 \mathrm{m} / \mathrm{s}\) through a \(3.6-\mathrm{T}\) magnetic field. The velocity \(\overrightarrow{\mathbf{v}}\) is perpendicular to the magnetic field. The electric field within the capacitor has a value of \(170 \mathrm{N} / \mathrm{C},\) and each plate has an area of \(7.5 \times 10^{-4} \mathrm{m}^{2} .\) What is the magnetic force (magnitude and direction) exerted on the positive plate of the capacitor?
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