91Ó°ÊÓ

A small particle with positive charge \(q=+3.75 \times 10^{-4} \mathrm{C}\) and mass \(m=5.00 \times 10^{-5} \mathrm{~kg}\) is moving in a region of uniform electric and magnetic fields. The magnetic field is \(B=4.00 \mathrm{~T}\) in the \(+z\) -direction. The electric field is also in the \(+z\) -direction and has magnitude \(E=60.0 \mathrm{~N} / \mathrm{C}\). At time \(t=0\) the particle is on the \(y\) -axis at \(y=+1.00 \mathrm{~m}\) and has velocity \(v=30.0 \mathrm{~m} / \mathrm{s}\) in the \(+x\) -direction. Neglect gravity. (a) What are the \(x\) -, \(y\) and \(z\) -coordinates of the particle at \(t=0.0200 \mathrm{~s} ?\) (b) What is the speed of the particle at \(t=0.0200 \mathrm{~s} ?\)

Step by step solution

01

Determine the Acceleration

The force on the particle due to the electric field is given by \( F = qE \). Using Newton's second law \( F = ma \), the acceleration \( a \) of the particle can be obtained as \( a = F/m = qE/m = (3.75 \times 10^{-4} \cdot 60) / (5.00 \times 10^{-5}) = 450 m/s^2 \). The acceleration is in the positive \( z \) direction.

02

Determine the Radius of the Circular Path

The force on the particle due to the magnetic field is \( F = qvB \). This force provides the centripetal force \( F = mv^2 / r \) for the circular motion. Equating the two forces gives \( r = mv / (qB) = (5.00 \times 10^{-5} \cdot 30.0) / (3.75 \times 10^{-4} \cdot 4.00) = 0.1 m \). The radius of the circular path is 0.1 m.

03

Calculate the Position Coordinates

Using the equations of motion: \( x(t) = vt, y(t) = y0 + r(1 - cos(\omega t)), z(t) = z0 + ut + 0.5at^2 \). Insert the given and calculated values: \( x = 30.0 \cdot 0.0200 = 0.600 \) m, \( y = 1.00 + 0.1(1 - cos(2 \pi \cdot 30.0 \cdot 0.0200)) = 1.05 \) m, \( z = 0 + 0 \cdot 0.0200 + 0.5 \cdot 450 \cdot (0.0200)^2 = 0.0900 \) m.

04

Determine the Speed

The speed \( v \) at time \( t \) can be obtained from \( v = u + at \) where \( u \) is the initial speed. Inserting the given and calculated values: \( v = 0 + 450 \cdot 0.0200 = 9.0 \) m/s.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Electric Field

An electric field represents how an electrically charged particle interacts with space around it. It is a vector quantity, meaning it has both a magnitude and direction. In our exercise, the electric field has a magnitude of \(60.0 \mathrm{~N} / \mathrm{C}\) and directs in the positive \(z\)-direction.

• The presence of an electric field applies a force on the charged particle, \( F = qE \), where \( q \) is the charge of the particle and \( E \) is the electric field strength.
• This force results in an acceleration, calculated using Newton's second law \( F = ma \), leading to \( a = qE/m \).

The interaction with the electric field makes the particle accelerate in its direction, affecting its motion.

Magnetic Field

A magnetic field exerts a force on a moving charged particle, which is perpendicular to both the velocity of the particle and the direction of the field itself. In our case, the magnetic field is \(4.00 \mathrm{~T}\) in intensity and aligns with the positive \(z\)-direction. This is crucial as it moves the charged particle sideways, not affecting its speed along the original velocity vector, but acting as a centripetal force.

• The force due to the magnetic field is \( F = qvB \), where \( v \) is the velocity of the particle and \( B \) is the magnetic field strength.
• This force contributes to the circular motion of the particle by continually deflecting its path perpendicularly.

Importantly, this component does not work on the particle because it does not change the speed of the particle, only its direction.

Centripetal Force

The concept of centripetal force is central to understanding how a charged particle moves in a magnetic field in circular motion. A centripetal force is any force that keeps an object moving in a circular path. For our charged particle, the magnetic field provides this centripetal force that sustains the circular motion.

• The centripetal force needed for circular motion is calculated as \( F = \frac{mv^2}{r} \), where \( m \) is the mass, \( v \) is velocity, and \( r \) is the radius of the path.
• By equating the magnetic force \( qvB \) to the expression for centripetal force, you can find the radius \( r \) of the circular path.

This balance of forces is what causes the particle to loop in a circular path rather than a straight line under the magnetic influence.

Acceleration

Acceleration is the change in velocity of the particle over time. In this exercise, it is the result of the electric force. Since the electric field points in the positive \(z\)-direction, the charged particle also accelerates in that direction.

• Given by \( a = \frac{qE}{m} \), it shows how the particle speeds up or slows down in the presence of an electric field.
• In the solution, we calculated it to be \(450 \mathrm{~m/s^2}\), affecting the position of the particle along the \(z\)-axis over time.

Understanding acceleration is essential as it impacts where the particle will be over time, together with forces from both electric and magnetic fields.

Circular Motion

A remarkable outcome of the interaction between a charged particle and a magnetic field is circular motion. This motion arises because the magnetic force serves as the centripetal force, steering the particle in a circle.

• For the charged particle, the radius \( r \) of this path is defined by the balance between magnetic force \( qvB \) and the centripetal force \( \frac{mv^2}{r} \).
• Using \( r = \frac{mv}{qB} \), we find the particle in this scenario moves in a circle with radius \(0.1\) m.

This concept is vital for understanding how particles behave in devices such as cyclotrons and mass spectrometers. The interplay of these forces leads to predictable, periodic behavior ideal for various applications in physics and engineering.