Ch. 27 - Magnetism

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook

Giancoli Douglas 5th edition

Ch. 27 - Magnetism

Problem 23

Chapter 26, Problem 23

A 720-KeV (kinetic energy) proton enters a 0.20-T field, in a plane perpendicular to the field. What is the radius of its path? See Section 23–8.

Verified step by step guidance

Determine the relationship between the radius of the circular path and the forces acting on the proton. The proton moves in a circular path due to the magnetic force acting as the centripetal force. The equation for this is:

F

_c = F_B, where

F

_c = mv²r and

FB=qvB

Rearrange the equation to solve for the radius of the path,

r

r = \frac{mv}{qB}

, where

m

is the mass of the proton,

v

is its velocity,

q

is the charge of the proton, and

B

is the magnetic field strength.

Use the kinetic energy of the proton to find its velocity. The kinetic energy is given by

KE = \frac{1}{2} mv

². Rearrange to solve for

v

v = sqrt(\frac{2KE}{m})

. Substitute the given kinetic energy (720 keV, converted to joules) and the mass of the proton (

1.67 × 10

^-27 kg).

Substitute the values for

v

m

q

(charge of the proton,

1.6 × 10

^-19 C), and

B

(0.20 T) into the radius equation

r = \frac{mv}{qB}

Simplify the expression to calculate the radius of the proton's path. Ensure all units are consistent (e.g., joules for energy, kilograms for mass, teslas for magnetic field strength) before performing the calculation.

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Key Concepts

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

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion, calculated using the formula KE = 1/2 mv², where m is mass and v is velocity. In this context, the proton has a kinetic energy of 720 KeV, which can be converted to joules for further calculations. Understanding kinetic energy is crucial for determining the velocity of the proton as it enters the magnetic field.

Magnetic Force and Motion

When a charged particle, like a proton, moves through a magnetic field, it experiences a magnetic force that acts perpendicular to both its velocity and the magnetic field direction. This force causes the particle to move in a circular path. The radius of this path can be determined using the formula r = mv/(qB), where r is the radius, m is mass, v is velocity, q is charge, and B is the magnetic field strength.

Lorentz Force

The Lorentz force is the combined effect of electric and magnetic forces on a charged particle. In this scenario, the magnetic component is responsible for the circular motion of the proton in the magnetic field. The magnitude of the Lorentz force is given by F = qvB, which is essential for calculating the radius of the proton's path as it relates to the balance between the magnetic force and the centripetal force required for circular motion.

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Lorentz Transformations of Velocity

Related Practice

Textbook Question

How much work is required to rotate the current loop (Fig. 27–23) in a uniform magnetic field $\vec{B}$ from (a) θ = 0° ( $\vec{μ}$ ∣∣ $\vec{B}$ ) to θ = 180°, (b) θ = 90° to θ = -90°.

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Textbook Question

(II) A current-carrying circular loop of wire (radius r, current I) is partially immersed in a magnetic field of constant magnitude B₀ directed out of the page as shown in Fig. 27–43. Determine the net force on the loop due to the field in terms of θ₀. (Note that θ₀ points to the dashed line, above which B = 0.)

Textbook Question

A stiff wire 50.0 cm long is bent at a right angle in the middle. One section lies along the z axis and the other is along the line y = 2x in the xy plane. A current of 20.0 A flows in the wire—down the z axis and out the wire in the xy plane. The wire passes through a uniform magnetic field given by = (0.285î ) T. Determine the magnitude and direction of the total force on the wire.

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Textbook Question

For a particle of mass m and charge q moving in a circular path in a magnetic field B, (a) show that its kinetic energy is proportional to r², the square of the radius of curvature of its path. Show that its angular momentum is L=qBr² , around the center of the circle.

Textbook Question

(III) A curved wire, connecting two points a and b, lies in a plane perpendicular to a uniform magnetic field $\vec{B}$ and carries a current I. Show that the resultant magnetic force on the wire, no matter what its shape, is the same as that on a straight wire connecting the two points carrying the same current I. See Fig. 27–44.

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Textbook Question

A particle of charge q moves in a circular path of radius r in a uniform magnetic field $\vec{B}$ . If the magnitude of the magnetic field is doubled, and the kinetic energy of the particle remains constant, what happens to the angular momentum of the particle?

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