Problem 27 - University Physics with Modern Physics

Question

A proton \(\left(q=1.60 \times 10^{-19} \mathrm{C}, m=1.67 \times 10^{-27} \mathrm{kg}\right)\) moves in a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}=(0.500 \mathrm{T}) \hat{\boldsymbol{i}} .\) At \(t=0\) the proton has velocity components \(v_{x}=1.50 \times 10^{5} \mathrm{m} / \mathrm{s}, v_{y}=0,\) and \(v_{z}=2.00 \times 10^{5} \mathrm{m} / \mathrm{s}(\text { see Example } 27.4) .\) (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, \(\vec{E}=\left(+2.00 \times 10^{4} \mathrm{V} / \mathrm{m}\right) \hat{\imath}\) (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

The magnetic force is \(1.60 \times 10^{-14} \text{ N}\) in the \(-\hat{j}\) direction. The proton accelerates in the electric field direction, but the field does not affect the helical radius. The half-period displacement is \(1.64 \times 10^{-2} \text{ m}\).

Step 1: Calculate Magnetic Force

The magnetic force on a charged particle is given by the Lorentz force equation: \[ \mathbf{F}_B = q(\mathbf{v} \times \mathbf{B}) \]Given: \[ q = 1.60 \times 10^{-19} \text{ C}, \quad \mathbf{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k} = (1.50 \times 10^5 \hat{i} + 0 \hat{j} + 2.00 \times 10^5 \hat{k}) \text{ m/s}, \quad \mathbf{B} = 0.500 \hat{i} \text{ T} \]First, compute the cross product, \( \mathbf{v} \times \mathbf{B} \):\[ \mathbf{v} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1.50 \times 10^5 & 0 & 2.00 \times 10^5 \ 0.5 & 0 & 0 \end{vmatrix} = (0 - 0)\hat{i} - (0 - (0.5 \cdot 2.00 \times 10^5))\hat{j} + (0 - 0)\hat{k} \]\[ = -1.00 \times 10^5 \hat{j} \text{ m/s} \text{ T} \]Now, find the magnetic force:\[ \mathbf{F}_B = 1.60 \times 10^{-19} \times (-1.00 \times 10^5) \hat{j} = -1.60 \times 10^{-14} \hat{j} \text{ N} \]Thus, the magnetic force has a magnitude of \(1.60 \times 10^{-14} \text{ N} \) in the \(-\hat{j}\) direction.

Step 2: Determine Effect of Electric Field

The electric force acting on the proton is given by:\[ \mathbf{F}_E = q\mathbf{E} \]Where:\( q = 1.60 \times 10^{-19} \text{ C} \) and \( \mathbf{E} = 2.00 \times 10^4 \hat{i} \text{ V/m} \)Thus:\[ \mathbf{F}_E = 1.60 \times 10^{-19} \times 2.00 \times 10^4 \hat{i} = 3.20 \times 10^{-15} \hat{i} \text{ N} \]Since the electric field is in the \( \hat{i} \) direction, it will accelerate the proton in the same direction. This results in an acceleration component in the direction of the electric field.

Step 3: Analyze Proton's Path and Impact of Electric Field

Under magnetic forces alone, the proton would describe a circular or helical path. The presence of a uniform electric field will add a drift velocity in the direction of this field.The radius of the helical path is determined by:\[ r = \frac{mv_{\perp}}{qB} \]Given the velocity components and the magnetic field, the perpendicular velocity \(v_{\perp} = v_z = 2.00 \times 10^5 \text{ m/s} \).Thus, the radius \( r \) is:\[ r = \frac{1.67 \times 10^{-27} \times 2.00 \times 10^5}{1.60 \times 10^{-19} \times 0.5} = 4.18 \times 10^{-3} \text{ m} \]The electric field does not affect the radius of the helical path, which is only influenced by non-parallel components of velocity relative to \( \mathbf{B} \).

Step 4: Compute Displacement at Half Period

The period \( T \) for circular motion is:\[ T = \frac{2\pi m}{qB} \]Given:\[ T = \frac{2\pi \times 1.67 \times 10^{-27}}{1.60 \times 10^{-19} \times 0.5} = 1.31 \times 10^{-7} \text{ s} \]At \( t = T/2 \), the proton moves halfway along the circle's chip due to the electric field; it will have moved a drift distance \( d \) in the \( x \)-direction:\[ d = \frac{F_E}{m} \left(\frac{T}{2}\right)^2 \]\[ = \frac{3.20 \times 10^{-15}}{1.67 \times 10^{-27}} \left(\frac{1.31 \times 10^{-7}}{2}\right)^2 \]\[ = 1.64 \times 10^{-2} \text{ m} \]Therefore, the \(x\)-component of displacement at \(t = T/2\) is approximately \(1.64 \times 10^{-2}\) m.

Key Concepts

Lorentz force

When analyzing the motion of charged particles, the Lorentz force equation is essential. It describes how charged particles experience force when they move through both electric and magnetic fields. In the presence of only a magnetic field, the Lorentz force is expressed by:

• \( \mathbf{F}_B = q(\mathbf{v} \times \mathbf{B}) \)

Here, \( q \) is the charge, \( \mathbf{v} \) is the velocity vector, and \( \mathbf{B} \) is the magnetic field vector.
The cross product \( \mathbf{v} \times \mathbf{B} \) results in a vector that is perpendicular to both \( \mathbf{v} \) and \( \mathbf{B} \). Thus, the magnetic force is always perpendicular to the particle's velocity and does no work on the particle.
This perpendicular force causes the charged particle to move in a circular or helical path, which is a key feature of magnetic interactions.

electric field effect on motion

The presence of an electric field introduces an additional force on charged particles, affecting their motion distinctly from a magnetic field. The electric force \( \mathbf{F}_E \) exerted on a charge \( q \) by an electric field \( \mathbf{E} \) is given by:

• \( \mathbf{F}_E = q\mathbf{E} \)

The electric force acts in the direction of the electric field \( \mathbf{E} \) and causes acceleration, changing the velocity of the particle linearly. Unlike the magnetic force, it can increase or decrease the kinetic energy of the proton.
In our problem, the electric field is aligned with the \( +x \)-axis, adding an acceleration component in this direction. As a result, the proton experiences both circular (due to magnetic force) and linear motion (due to the electric field), resulting in a helical trajectory.

helical motion of charged particles

Charged particles display helical motion when subject to both perpendicular magnetic and electric fields. In such scenarios, the interaction of forces results in a unique path.
Initially, a charged particle moving perpendicular to a magnetic field traces a circular path because the Lorentz force only changes the direction of velocity without affecting its magnitude. This circular motion occurs in a plane orthogonal to the magnetic field.
Adding an electric field parallel to one of the velocity components results in helical motion. The circle's radius, determined by \( \frac{mv_{\perp}}{qB} \), remains unaffected by the electric field. However, the electric field imparts a drift motion in its direction, producing the helical shape, which is a combination of linear and circular motion.

cross product in physics

In physics, the cross product is a mathematical operation widely used to find the resultant vector perpendicular to the two original vectors. For vectors \( \mathbf{a} \) and \( \mathbf{b} \), their cross product \( \mathbf{a} \times \mathbf{b} \) can be computed using the determinant of a matrix:

• \( \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ a_{x} & a_{y} & a_{z} \ b_{x} & b_{y} & b_{z} \end{vmatrix} \)

This operation gives a vector orthogonal to both \( \mathbf{a} \) and \( \mathbf{b} \).
In our context, the cross product \( \mathbf{v} \times \mathbf{B} \) is vital because it determines the direction and magnitude of the magnetic force in the Lorentz force calculation. This highlights why understanding vector products is crucial for analyzing forces and motions in magnetic and electric fields.