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Ch 27: Magnetic Field and Magnetic Forces
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 14th EditionCh 27: Magnetic Field and Magnetic ForcesProblem 21
Chapter 27, Problem 21

A deuteron (the nucleus of an isotope of hydrogen) has a mass of 3.34 x 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?

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1
To find the speed of the deuteron, we use the formula for the radius of the circular path in a magnetic field: \( r = \frac{mv}{qB} \). Here, \( r \) is the radius, \( m \) is the mass of the deuteron, \( v \) is the speed, \( q \) is the charge, and \( B \) is the magnetic field. Rearrange the formula to solve for \( v \): \( v = \frac{qBr}{m} \). Substitute the given values: \( q = e = 1.60 \times 10^{-19} \) C, \( B = 2.50 \) T, \( r = 6.96 \times 10^{-3} \) m, and \( m = 3.34 \times 10^{-27} \) kg.
To find the time required for the deuteron to make half a revolution, first calculate the circumference of the circular path: \( C = 2\pi r \). Since we need the time for half a revolution, use \( \frac{C}{2} \). The time \( t \) can be found using \( t = \frac{\text{distance}}{\text{speed}} \). Substitute the values for the distance (half the circumference) and the speed calculated in step 1.
To find the potential difference required to accelerate the deuteron to the calculated speed, use the kinetic energy formula \( KE = \frac{1}{2}mv^2 \). The work-energy principle states that the work done on the deuteron is equal to its kinetic energy, and this work is given by \( qV \), where \( V \) is the potential difference. Set \( qV = \frac{1}{2}mv^2 \) and solve for \( V \): \( V = \frac{mv^2}{2q} \). Substitute the values for \( m \), \( v \), and \( q \).
Ensure all units are consistent when substituting values into the equations. Convert the radius from millimeters to meters, and verify that the charge is in coulombs.
Review the concepts of circular motion in a magnetic field, kinetic energy, and potential difference to understand the physical principles behind each calculation. This will help reinforce the understanding of how magnetic fields influence charged particles and how energy transformations occur.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Magnetic Force on a Charged Particle

When a charged particle moves through a magnetic field, it experiences a force perpendicular to both its velocity and the magnetic field direction. This force causes the particle to move in a circular path, and its magnitude is given by F = qvB, where q is the charge, v is the velocity, and B is the magnetic field strength.
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Centripetal Force and Circular Motion

For a particle moving in a circular path, the centripetal force required to maintain this motion is provided by the magnetic force. The centripetal force is given by F = mv^2/r, where m is the mass, v is the velocity, and r is the radius of the circular path. Equating this to the magnetic force allows us to solve for the particle's speed.
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Energy and Potential Difference

The kinetic energy acquired by a charged particle when accelerated through a potential difference is given by KE = qV, where q is the charge and V is the potential difference. This energy is also expressed as KE = 0.5mv^2, allowing us to find the potential difference needed to achieve a specific speed by equating these expressions.
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Related Practice
Textbook Question

In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm. You are redesigning the cyclotron to be used instead for alpha particles with energy 300 keV. An alpha particle has charge q = +2e and mass m = 6.64 x 10-27 kg. If the magnetic field isn't changed, what will be the orbital radius of the alpha particles?

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Textbook Question

A 150 V battery is connected across two parallel metal plates of area 28.5 cm2 and separation 8.20 mm. A beam of alpha particles (charge +2e, mass 6.64 x 10-27 kg) is accelerated from rest through a potential difference of 1.75 kV and enters the region between the plates perpendicular to the electric field, as shown in Fig. E27.29. What magnitude and direction of magnetic field are needed so that the alpha particles emerge undeflected from between the plates?

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Textbook Question

Cyclotrons are widely used in nuclear medicine for producing short-lived radioactive isotopes. These cyclotrons typically accelerate H- (the hydride ion, which has one proton and two electrons) to an energy of 5 MeV to 20 MeV. This ion has a mass very close to that of a proton because the electron mass is negligible — about 1/2000 of the proton's mass. A typical magnetic field in such cyclotrons is 1.9 T. (a) What is the speed of a 5.0 MeV H-? (b) If the H- has energy 5.0 MeV and B = 1.9 T, what is the radius of this ion's circular orbit?

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Textbook Question

An open plastic soda bottle with an opening diameter of 2.5 cm is placed on a table. A uniform 1.75 T magnetic field directed upward and oriented 25° from vertical encompasses the bottle. What is the total magnetic flux through the plastic of the soda bottle?

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Textbook Question

A beam of protons traveling at 1.20 km/s enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field, leaving the field in a direction perpendicular to its original direction (Fig. E27.24) . The beam travels a distance of 1.18 cm while in the field. What is the magnitude of the magnetic field?

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Textbook Question

A 150 g ball containing 4.00 x 108 excess electrons is dropped into a 125 m vertical shaft. At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has magnitude 0.250 T and direction from east to west. If air resistance is negligibly small, find the magnitude and direction of the force that this magnetic field exerts on the ball just as it enters the field.

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