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Chapter 27: Problem 20

A deuteron (the nucleus of an isotope of hydrogen) has a mass of \(3.34 \times 10^{-27} \mathrm{~kg}\) and a charge of \(+e .\) The deuteron travels in a circular path with a radius of \(6.96 \mathrm{~mm}\) in a magnetic field with magnitude \(2.50 \mathrm{~T}\). (a) Find the speed of the deuteron. (b) Find the time required for it to make half a revolution. (c) Through what potential difference would the deuteron have to be accelerated to acquire this speed?

Short Answer

Expert verified
After solving the above steps, you should find: (a) The speed of the deuteron, (b) The time required for the deuteron to make half a revolution, and (c) The potential difference the deuteron must be accelerated through.

Step by step solution

01

Calculate the Velocity

From physics, we know that the magnetic force acting on a moving charged particle in a magnetic field can be described by the equation \(F = qvB\sin{\theta}\), where \(q\) is the charge, \(v\) is the velocity, \(B\) is the magnetic field strength, and \(\theta\) is the angle between the velocity and the magnetic field vectors. For circular motion in a magnetic field, the magnetic force is the centripetal force. Because the magnetic field is perpendicular to the motion of the deuteron (i.e., \(\sin{\theta} = 1\)), the equation simplifies to \(qvB = mv^2/r\), where \(m\) is the mass and \(r\) is the radius of the circular path. Solving for \(v\), we get \(v = qrB/m\). Substituting the given values, we can calculate the speed of the deuteron.
02

Calculate the Time for Half Revolution

The period of revolution, \(T\), is the time taken for one complete revolution, which is given by the equation \(T = 2\pi r/v\). Since we're asked for the time it takes for half a revolution, we simply divide \(T\) by 2 to get \(T/2 = \pi r/v\). By substituting the already calculated speed of the deuteron, we can find the required time.
03

Find the Potential Difference

The potential difference through which the deuteron must be accelerated for it to acquire the calculated speed can be found by applying the energy conservation principle. The kinetic energy gained by the deuteron equals the work done by the electric field force, which is described by '\(\Delta U = 1/2 mv^2 = q\Delta V\)', where \(\Delta V\) is the potential difference. Solving for \(\Delta V\), we find \(\Delta V = 1/2 (mv^2)/q\). Substituting the calculated speed and given mass and charge, we will find the required potential difference.

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Key Concepts

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

Circular Motion
Circular motion occurs when an object moves along a circular path. It involves constant change in direction, which means even if the object maintains constant speed, its velocity is not constant because velocity includes direction. For charged particles, like a deuteron moving in a magnetic field, this circular path is due to the Lorentz force influencing its trajectory.
In circular motion, the centripetal force is responsible for keeping the object on its path. Specifically for particles in a magnetic field, this force arises due to the interaction between the magnetic field and the charge. The particle's charge enters perpendicular to the magnetic field, inducing this circular motion. The equation related to circular motion in a magnetic field simplifies as:
  • The magnetic force equals the centripetal force: \[ F = qvB = \frac{mv^2}{r} \]
  • This relationship allows us to calculate the velocity.
Understanding these interactions allows us to predict how particle speed and circular path radius relate under magnetic influence.
Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. For any object moving at velocity, its kinetic energy can be calculated using the equation:
  • \[ KE = \frac{1}{2} mv^2 \]
  • Here, \(m\) represents the mass and \(v\) is the velocity.
For the deuteron in our example, its speed, calculated from its circular motion in the magnetic field, is crucial to determining its kinetic energy. The kinetic energy gives us insights into how much work the particle can perform due to its motion.
When a charged particle is accelerated in an electric field, its increase in kinetic energy is directly related to the work done by the electric potential. Hence, this energy change is vital because it correlates with the potential difference applied to accelerate the particle initially.
Potential Difference
Potential difference, often expressed in volts, signifies the work needed to move a charge between two points. In physics, it represents the energy supplied per charge by an electric field. In our deuteron scenario, the potential difference is central to accelerating it to a specified speed.
Using energy conservation principles, we equate the kinetic energy with the work done by the electric field. This yields:
  • \[ q\Delta V = \frac{1}{2} mv^2 \]
  • Solving for \(\Delta V\), we have: \[ \Delta V = \frac{1}{2} \left(\frac{mv^2}{q}\right) \]
  • \(\Delta V\) reveals how much electric potential is necessary to accelerate the deuteron to its velocity.
Thus, potential difference measures the effectiveness of an electric field in increasing a particle鈥檚 energy and speed.
Centripetal Force
Centripetal force is crucial for maintaining circular motion. This force acts perpendicular to the velocity direction, aiming towards the circle's center, and is necessary for changing the direction of velocity without altering speed. In our scenario, for a deuteron moving in a magnetic field, the magnetic force itself provides this necessary centripetal force.
As derived in circular motion:
  • \[ F_c = \frac{mv^2}{r} = qvB \]
  • Centripetal force must equal magnetic force for the deuteron to maintain its circular path.
These forces being equal ensure that any variations in the magnetic field or deuteron's speed directly impact the radius and speed of the circular motion.
Essentially, understanding centripetal force helps us determine the relationship between force, velocity, and radius in the presence of a magnetic field.

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Most popular questions from this chapter

A wire \(25.0 \mathrm{~cm}\) long lies along the \(z\) -axis and carries a current of \(7.40 \mathrm{~A}\) in the \(+z\) -direction. The magnetic field is uniform and has components \(B_{x}=-0.242 \mathrm{~T}, B_{y}=-0.985 \mathrm{~T},\) and \(B_{z}=-0.336 \mathrm{~T}\) (a) Find the components of the magnetic force on the wire. (b) What is the magnitude of the net magnetic force on the wire?

A uniform bar of length \(L\) carries a current \(I\) in the direction from point \(a\) to point \(b\) (Fig. \(\mathbf{P 2 7 . 6 6}\) ). The bar is in a uniform magnetic field that is directed into the page. Consider the torque about an axis perpendicular to the bar at point \(a\) that is due to the force that the magnetic field exerts on the bar. (a) Suppose that an infinitesimal section of the bar has length \(d x\) and is located a distance \(x\) from point \(a\). Calculate the torque \(d \tau\) about point \(a\) due to the magnetic force on this infinitesimal section. (b) Use \(\tau=\int_{a}^{b} d \tau\) to calculate the total torque \(\tau\) on the bar. (c) Show that \(\tau\) is the same as though all of the magnetic force acted at the midpoint of the bar.

A particle with initial velocity \(\overrightarrow{\boldsymbol{v}}_{0}=\left(5.85 \times 10^{3} \mathrm{~m} / \mathrm{s}\right) \hat{\jmath}\) enters a region of uniform electric and magnetic fields. The magnetic field in the region is \(\overrightarrow{\boldsymbol{B}}=-(1.35 \mathrm{~T}) \hat{\boldsymbol{k}} .\) Calculate the magnitude and direction of the electric field in the region if the particle is to pass through undeflected, for a particle of charge (a) \(+0.640 \mathrm{nC}\) and (b) \(-0.320 \mathrm{nC}\). You can ignore the weight of the particle.

A circular area with a radius of \(6.50 \mathrm{~cm}\) lies in the \(x y\) -plane. What is the magnitude of the magnetic flux through this circle due to a uniform magnetic field \(B=0.230 \mathrm{~T}\) (a) in the \(+z\) -direction; (b) at an angle of \(53.1^{\circ}\) from the \(+z\) -direction; (c) in the \(+y\) -direction?

You are using a type of mass spectrometer to measure charge-to-mass ratios of atomic ions. In the device, atoms are ionized with a beam of electrons to produce positive ions, which are then accelerated through a potential difference \(V\). (The final speed of the ions is great enough that you can ignore their initial speed.) The ions then enter a region in which a uniform magnetic field \(\vec{B}\) is perpendicular to the velocity of the ions and has magnitude \(B=0.250 \mathrm{~T}\). In this \(\overrightarrow{\boldsymbol{B}}\) region, the ions move in a semicircular path of radius \(R .\) You measure \(R\) as a function of the accelerating voltage \(V\) for one particular atomic ion: $$ \begin{array}{l|lllll} \boldsymbol{V}(\mathbf{k} \mathbf{V}) & 10.0 & 12.0 & 14.0 & 16.0 & 18.0 \\ \hline \boldsymbol{R}(\mathrm{cm}) & 19.9 & 21.8 & 23.6 & 25.2 & 26.8 \end{array} $$ (a) How can you plot the data points so that they will fall close to a straight line? Explain. (b) Construct the graph described in part (a). Use the slope of the best-fit straight line to calculate the charge-to-mass ratio \((q / m)\) for the ion. \((\mathrm{c})\) For \(V=20.0 \mathrm{kV},\) what is the speed of the ions as they enter the \(\vec{B}\) region? (d) If ions that have \(R=21.2 \mathrm{~cm}\) for \(V=12.0 \mathrm{kV}\) are singly ionized, what is \(R\) when \(V=12.0 \mathrm{kV}\) for ions that are doubly ionized?
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