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Ch. 27 - Magnetism
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoGiancoli Douglas 5th editionCh. 27 - MagnetismProblema 76
Capitolo 26, Problema 76

Suppose the Earth’s magnetic field at the equator has magnitude 0.50 x 10⁻⁴ T and a northerly direction at all points. Estimate the speed a singly ionized uranium ion ( m = 238 u, q = e) would need to circle the Earth 5.0 km above the equator. Can you ignore gravity? [Ignore relativity.]

Guida verificata passo dopo passo
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Step 1: Recognize that the problem involves a charged particle (a singly ionized uranium ion) moving in a circular path due to the magnetic force exerted by the Earth's magnetic field. The magnetic force provides the centripetal force required for circular motion.
Step 2: Write the equation for the centripetal force: \( F_c = \frac{m v^2}{r} \), where \( m \) is the mass of the ion, \( v \) is its speed, and \( r \) is the radius of the circular path. The radius \( r \) is the Earth's radius plus the height above the equator (\( r = R_E + h \)).
Step 3: Write the equation for the magnetic force: \( F_B = q v B \), where \( q \) is the charge of the ion, \( v \) is its speed, and \( B \) is the magnetic field strength. Since the magnetic force provides the centripetal force, set \( F_B = F_c \): \( q v B = \frac{m v^2}{r} \).
Step 4: Simplify the equation to solve for the speed \( v \): \( v = \frac{q B r}{m} \). Substitute \( r = R_E + h \), where \( R_E \) is the Earth's radius (approximately \( 6.37 \times 10^6 \ \text{m} \)) and \( h = 5.0 \ \text{km} = 5.0 \times 10^3 \ \text{m} \).
Step 5: Substitute the known values into the equation: \( q = e = 1.6 \times 10^{-19} \ \text{C} \), \( B = 0.50 \times 10^{-4} \ \text{T} \), \( m = 238 \ \text{u} = 238 \times 1.66 \times 10^{-27} \ \text{kg} \), and \( r = R_E + h \). Perform the calculation to find \( v \). Note that gravity can be ignored because the magnetic force dominates in this scenario.

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Magnetic Force on a Charged Particle

A charged particle moving in a magnetic field experiences a magnetic force perpendicular to both its velocity and the magnetic field direction. This force can be calculated using the equation F = qvB sin(θ), where F is the magnetic force, q is the charge, v is the velocity, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field. For a particle moving perpendicular to the field, θ is 90 degrees, simplifying the equation to F = qvB.
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Magnetic Force Between Two Moving Charges

Centripetal Force

When an object moves in a circular path, it requires a centripetal force directed towards the center of the circle to maintain its motion. This force can be expressed as F_c = mv²/r, where m is the mass of the object, v is its speed, and r is the radius of the circular path. In this scenario, the magnetic force acting on the uranium ion provides the necessary centripetal force to keep it in circular motion around the Earth.
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Intro to Centripetal Forces

Radius of Circular Motion in a Magnetic Field

The radius of the circular path of a charged particle in a magnetic field can be determined by equating the magnetic force to the centripetal force. This leads to the formula r = mv/(qB), where r is the radius, m is the mass of the particle, v is its speed, q is its charge, and B is the magnetic field strength. In this problem, the radius is the distance from the center of the Earth to the ion's altitude above the equator, which is crucial for calculating the required speed.
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Circular Motion of Charges in Magnetic Fields
Pratica correlata
Domanda del libro di testo

Near the equator, the Earth’s magnetic field points almost horizontally to the north and has magnitude B = 0.50 x 10⁻⁴ T. What should be the magnitude and direction for the velocity of an electron if its weight is to be exactly balanced by the magnetic force?

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Domanda del libro di testo

A uniform conducting rod of length ℓ and mass m sits atop a fulcrum, which is placed a distance ℓ/4 from the rod’s left-hand end and is immersed in a uniform magnetic field of magnitude B directed into the page (Fig. 27–54). An object whose mass M is 7.0 times greater than the rod’s mass is hung from the rod’s left-hand end. What current (direction and magnitude) should flow through the rod in order for it to be “balanced” (i.e., be at rest horizontally) on the fulcrum? (Flexible connecting wires which exert negligible force on the rod are not shown.)


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The magnetic field B at the center of a circular coil of wire carrying a current I (as in Fig. 27–9) is B = (μ₀NI) / 2r where N is the number of loops in the coil and r is its radius. Imagine a simple model in which the Earth’s magnetic field of about 1 G ( = 1 x 10⁻⁴ T) near the poles is produced by a single current loop around the equator. Roughly estimate the current this loop would carry.

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