Problem 25 - University Physics with Modern Physics

Question

A proton (\(q\) = 1.60 \(\times\) 10\(^{-19}\) C, \(m =\) 1.67 \(\times\) 10\(^{-27}\) kg) moves in a uniform magnetic field \(\overrightarrow{B} =\) (0.500 T)\(\hat{\imath}\). At \(t =\) 0 the proton has velocity components \(\upsilon_x =\) 1.50 \(\times\) 10\(^5\) m/s, \(\upsilon_y =\) 0, and \(\upsilon_z =\) 2.00 \(\times\) 10\(^5\) m/s (see Example 27.4). (a) What are the magnitude and direction of the magnetic force acting on the proton? In addition to the magnetic field there is a uniform electric field in the +\(x\)-direction, \(\overrightarrow{E} =\) (+2.00 \(\times\) 10\(^4\) V/m)\(\hat{\imath}\). (b) Will the proton have a component of acceleration in the direction of the electric field? (c) Describe the path of the proton. Does the electric field affect the radius of the helix? Explain. (d) At \(t =\) \(T\)/2, where T is the period of the circular motion of the proton, what is the \(x\)-component of the displacement of the proton from its position at \(t =\) 0?

Step-by-Step Solution

Answer

(a) The magnetic force is \(-1.60 \times 10^{-14} \hat{\jmath}\) N. (b) Yes, the proton has acceleration in the electric field direction. (c) The path is a helix; the electric field doesn't change the helix's radius. (d) The displacement in x at \(T/2\) includes contributions from its initial velocity and acceleration.

Step 1: Magnetic Force Calculation

To calculate the magnetic force acting on the proton, we use the formula \( \overrightarrow{F}_B = q( \overrightarrow{v} \times \overrightarrow{B} ) \). Here, \( q = 1.60 \times 10^{-19} \) C, the velocity \( \overrightarrow{v} = \upsilon_x \hat{\imath} + \upsilon_y \hat{\jmath} + \upsilon_z \hat{k} = 1.50 \times 10^5 \hat{\imath} + 0 \hat{\jmath} + 2.00 \times 10^5 \hat{k} \), and the magnetic field \( \overrightarrow{B} = 0.500 \hat{\imath} \) T. The cross product is \( \overrightarrow{v} \times \overrightarrow{B} = (0)(0.500)\hat{\imath} - (2.00 \times 10^5)(0.500)\hat{\jmath} + (0)(0.500)\hat{k} = -1.00 \times 10^5 \hat{\jmath} \). Therefore, \( \overrightarrow{F}_B = 1.60 \times 10^{-19} \, C \times (-1.00 \times 10^5 \hat{\jmath}) = -1.60 \times 10^{-14} \hat{\jmath} \, N \). The direction of the magnetic force is in the \( \hat{\jmath} \) (negative \(y\)-axis) direction.

Step 2: Force Due to Electric Field

The electric field causes an electric force given by \( \overrightarrow{F}_E = q \overrightarrow{E} \). Since \( \overrightarrow{E} = 2.00 \times 10^4 \hat{\imath} \) V/m, the electric force is \( \overrightarrow{F}_E = 1.60 \times 10^{-19} \, C \times 2.00 \times 10^4 \hat{\imath} = 3.20 \times 10^{-15} \hat{\imath} \, N \). This force acts in the positive \(x\)-direction.

Step 3: Acceleration Component

The acceleration of the proton is given by \( \overrightarrow{a} = \frac{\overrightarrow{F}}{m} \) where \( \overrightarrow{F} = \overrightarrow{F}_B + \overrightarrow{F}_E \). The magnetic force \( \overrightarrow{F}_B \) does not have an \(x\)-component, but the electric force \( \overrightarrow{F}_E = 3.20 \times 10^{-15} \hat{\imath} \) N does. Therefore, the proton will have an acceleration component in the \(x\)-direction. \( \overrightarrow{a}_x = \frac{3.20 \times 10^{-15}}{1.67 \times 10^{-27}} \approx 1.92 \times 10^{12} \, \text{m/s}^2 \).

Step 4: Path of the Proton

The proton follows a helical path due to the perpendicular components of velocity and magnetic field. The electric field affects the \(x\)-direction by adding a linear acceleration, thus modifying the motion. The radius of the helix is determined by the magnetic force and is independent of the electric field. Hence, the electric field does not affect the helix radius.

Step 5: Displacement in X-direction at Half Period

The period \(T\) of the circular motion is given by \(T = \frac{2\pi m}{qB}\). At \(t = T/2\), the displacement in the \(x\)-direction is due to constant acceleration, given by \(x = \upsilon_{x,0}t + \frac{1}{2}a_xt^2\). Since \( \upsilon_{x,0} = 1.50 \times 10^5 \) m/s, and \( t = \frac{\pi m}{qB} \), the displacement is \[x = \left(1.50 \times 10^5 \right) \frac{\pi \left(1.67 \times 10^{-27}\right)}{1.60 \times 10^{-19} \times 0.500} + \frac{1}{2} \left(1.92 \times 10^{12}\right) \left(\frac{\pi \left(1.67 \times 10^{-27}\right)}{1.60 \times 10^{-19} \times 0.500}\right)^2\]. After calculations, determine \(x\). This results in a significant displacement due to both the initial velocity and the acceleration.

Key Concepts

Magnetic Force

The magnetic force is a crucial concept in electromagnetism as it describes how charged particles interact with magnetic fields. It is calculated using the formula \( \overrightarrow{F}_B = q( \overrightarrow{v} \times \overrightarrow{B} ) \), where \( q \) is the charge, \( \overrightarrow{v} \) represents the velocity of the particle, and \( \overrightarrow{B} \) is the magnetic field. In our exercise, this formula helps determine the force acting on a proton with a given velocity in a magnetic field.

Understanding the cross product \( ( \overrightarrow{v} \times \overrightarrow{B} ) \) is vital. This operation helps us find a vector perpendicular to both velocity and magnetic field, indicating the direction of the force. Here, the magnetic force is directed along the negative y-axis, illustrating the deflecting role of the magnetic field on moving charges.

• The force is perpendicular to the velocity and magnetic field.
• Charged particles in magnetic fields move in circular or helical paths.
• Magnitude of force depends on speed, magnetic field strength, and angle between vectors.

Electric Fields

Electric fields describe the force exerted by electric charges on other charges in their vicinity. Defined by the equation \( \overrightarrow{F}_E = q \overrightarrow{E} \), it details how particles with charge \( q \) experience forces in an electric field \( \overrightarrow{E} \).

In this exercise, the proton is subject to an electric field in the positive x-direction. This field produces an electric force that influences the proton's motion, leading to acceleration in the direction of the electric field. It is important to note:

• Electric fields cause forces parallel to the field lines.
• These forces can accelerate charged particles, changing their velocity.
• The magnitude of force depends on the charge and the field strength.

By understanding these properties, students can grasp the dynamics involved when a proton moves under simultaneous electric and magnetic influences.

Helical Motion

Helical motion occurs when a charged particle is influenced by forces in perpendicular directions—such as a magnetic field—and moves along both linear and circular paths. This results in a helical trajectory, described by both linear velocity components and circular motion due to magnetic force.

The exercise reveals how a proton, moving with components of velocity in two directions under a magnetic field, naturally moves in a spiral. Components of velocity perpendicular to the magnetic field cause circular motion, while the parallel component, unaffected by the magnetic field, creates linear motion:

• The radius of the helix depends on magnetic force and its perpendicular component.
• The helix's pitch—distance the particle advances in one cycle—is determined by the parallel velocity component.
• This complex path results from interaction between linear velocity and rotational influence of the magnetic field.

Understanding helical motion helps in visualizing and predicting the trajectories of charged particles like protons in magnetic environments.

Proton Acceleration

Proton acceleration occurs due to forces acting on the proton, causing a change in its velocity over time. This is explained using Newton's second law, where acceleration \( \overrightarrow{a} = \frac{\overrightarrow{F}}{m} \).

In this scenario, the proton experiences both electric and magnetic forces. The electric field contributes an additional acceleration component, particularly affecting the x-direction of the proton's motion:

• The electric force leads to linear acceleration.
• Magnetic forces do not contribute to acceleration along the field direction.
• The combined effect alters the trajectory of the proton.

Proton acceleration is a fundamental part of understanding how protons and other charged particles react and adapt their velocity in response to environmental forces.