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Chapter 27: Problem 21

A deuteron (the nucleus of an isotope of hydrogen) has a mass of 3.34 \(\times\) 10\(^{-27}\) kg and a charge of \(+e\). The deuteron travels in a circular path with a radius of 6.96 mm in a magnetic field with magnitude 2.50 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
The speed is about 3.03 \times 10^{6} m/s. Half-revolution time is about 7.2 \times 10^{-9} s. Potential difference is about 47.3 kV.

Step by step solution

01

Identify the Relevant Formula for Speed

To find the speed of the deuteron when it travels in a magnetic field, use the formula for centripetal force: \ F_c = qvB = \frac{mv^2}{r} \ where \( q \) is the charge of the deuteron, \( v \) is the velocity, \( B \) is the magnetic field, \( m \) is the mass, and \( r \) is the radius. Rearranging gives: \ v = \frac{qBr}{m}.
02

Calculate the Speed of the Deuteron

Substitute the given values into the equation: \ q = 1.6 \times 10^{-19} \text{C} (charge of deuteron) \ B = 2.50 \text{T}, \ r = 6.96 \times 10^{-3} \text{m} (convert mm to m), \ m = 3.34 \times 10^{-27} \text{kg}. \ v = \frac{(1.6 \times 10^{-19} \text{C})(2.50 \text{T})(6.96 \times 10^{-3} \text{m})}{3.34 \times 10^{-27} \text{kg}} \ Calculate to find \( v \).
03

Calculate Half Revolution Time

The time for half a revolution is found by calculating the period \( T \) and halving it. The period is given by \( T = \frac{2\pi r}{v} \). Calculate this period and then divide by 2 to get the time for half a revolution.
04

Calculate Potential Difference Required

The kinetic energy gained by the deuteron when accelerated through a potential difference \( V \) can be expressed as: \ KE = \frac{1}{2}mv^2 = qV \ Solve for \( V \) using \( v \) from Step 2: \ V = \frac{\frac{1}{2}mv^2}{q}.
05

Solve for the Potential Difference

Plug in values: \ m = 3.34 \times 10^{-27} \text{kg}, \ v = \text{value from Step 2}, \ q = 1.6 \times 10^{-19} \text{C}. \ Calculate \( V \) to find the potential difference.

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Key Concepts

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

Magnetic Field
In physics, a magnetic field is an invisible field that exerts a force on particles that are charged or moving through it. Magnetic fields surround magnets, electric currents, and changing electric fields. For a charging particle like a deuteron, when it enters a magnetic field, it experiences a force that is perpendicular to both the velocity of the particle and the direction of the magnetic field. This force can be described by the equation \( F = qvB \), where \( F \) is the magnetic force, \( q \) is the charge of the particle, \( v \) is the velocity, and \( B \) is the magnetic field strength.
  • This perpendicular force causes the particle to move in a circular path.
  • The strength of the magnetic field affects the radius of this circular motion.
  • The direction of the magnetic field determines which way the particle will curve.
Centripetal Force
Centripetal force is the force needed to make an object move in a circle. For a deuteron moving in a magnetic field, this force is provided by the magnetic force. The centripetal force ensures that the deuteron remains in circular motion. The formula for centripetal force \( F_c \) is given as \( F_c = \frac{mv^2}{r} \), where \( m \) is the mass of the deuteron, \( v \) is the velocity, and \( r \) is the radius of the circular path.
  • The force keeps the particle moving in its circular path and prevents it from flying off tangentially.
  • It is directed towards the center of the circle along which the deuteron is moving.
  • The magnetic force acting on the deuteron provides the necessary centripetal force.
Potential Difference
Potential difference, also known as voltage, is the energy required to move a charge between two points in a field. For a deuteron, to achieve a specific speed, it needs to be accelerated through a potential difference. This is expressed in the relation between kinetic energy and potential difference:\[ KE = \frac{1}{2}mv^2 = qV \]Here, \( KE \) is kinetic energy, \( m \) is mass, \( v \) is velocity, \( q \) is charge, and \( V \) is potential difference.
  • Potential difference gives the deuteron the energy needed to obtain its speed in the magnetic field.
  • Voltage is crucial for accelerating the particle to a speed where magnetic and centripetal forces balance.
  • This relationship shows the direct link between energy, speed, and voltage.
Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. For the deuteron, once it travels through a potential difference, it will gain kinetic energy proportionate to its speed. The equation for kinetic energy is given by:\[ KE = \frac{1}{2}mv^2 \]Kinetic energy depends on both the mass of the particle and the square of its velocity.
  • A faster-moving deuteron has higher kinetic energy.
  • This energy depends on the speed achieved when the deuteron is accelerated through a potential difference.
  • Kinetic energy demonstrates how fast the particle is able to move once it enters the magnetic field.

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Most popular questions from this chapter

A straight, 2.5-m wire carries a typical household current of 1.5 A (in one direction) at a location where the earth's magnetic field is 0.55 gauss from south to north. Find the magnitude and direction of the force that our planet's magnetic field exerts on this wire if it is oriented so that the current in it is running (a) from west to east, (b) vertically upward, (c) from north to south. (d) Is the magnetic force ever large enough to cause significant effects under normal household conditions?

An electromagnet produces a magnetic field of 0.550 T in a cylindrical region of radius 2.50 cm between its poles. A straight wire carrying a current of 10.8 A passes through the center of this region and is perpendicular to both the axis of the cylindrical region and the magnetic field. What magnitude of force does this field exert on the wire?

A conducting bar with mass m and length \(L\) slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current \(I\) in the rails and bar, and a constant, uniform, vertical magnetic field \(\overrightarrow{B}\) fills the region between the rails (\(\textbf{Fig. P27.65}\)). (a) Find the magnitude and direction of the net force on the conducting bar. Ignore friction, air resistance, and electrical resistance. (b) If the bar has mass \(m\), find the distance \(d\) that the bar must move along the rails from rest to attain speed \(v\). (c) It has been suggested that rail guns based on this principle could accelerate payloads into earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the earth (11.2 km/s). Let \(B =\) 0.80 T, \(I =\) 2.0 \(\times\) 10\(^3\) A, \(m =\) 25 kg, and \(L =\) 50 cm. For simplicity assume the net force on the object is equal to the magnetic force, as in parts (a) and (b), even though gravity plays an important role in an actual launch in space.

A horizontal rectangular surface has dimensions 2.80 cm by 3.20 cm and is in a uniform magnetic field that is directed at an angle of 30.0\(^\circ\) above the horizontal. What must the magnitude of the magnetic field be to produce a flux of 3.10 \(\times\) 10\(^{-4}\) Wb through the surface?

A mass spectrograph is used to measure the masses of ions, or to separate ions of different masses (see Section 27.5). In one design for such an instrument, ions with mass \(m\) and charge \(q\) are accelerated through a potential difference \(V\). They then enter a uniform magnetic field that is perpendicular to their velocity, and they are deflected in a semicircular path of radius \(R\). A detector measures where the ions complete the semicircle and from this it is easy to calculate \(R\). (a) Derive the equation for calculating the mass of the ion from measurements of \(B\), \(V\), \(R\), and \(q\). (b) What potential difference \(V\) is needed so that singly ionized \(^{12}\)C atoms will have \(R =\) 50.0 cm in a 0.150-T magnetic field? (c) Suppose the beam consists of a mixture of \(^{12}\)C and \(^{14}\)C ions. If \(v\) and \(B\) have the same values as in part (b), calculate the separation of these two isotopes at the detector. Do you think that this beam separation is sufficient for the two ions to be distinguished? (Make the assumption described in Problem 27.59 for the masses of the ions.)
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