Ch 27: Magnetic Field and Magnetic Forces

Young & Freedman Calc - University Physics 14th Edition

Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Non è quello che usi tu?Cambia libro di testo

Young & Freedman Calc 14th Edition

Ch 27: Magnetic Field and Magnetic Forces

Problema 29

Capitolo 27, Problema 29

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?

Guida verificata passo dopo passo

First, understand that for the alpha particles to emerge undeflected, the magnetic force must balance the electric force acting on them. The electric force \( F_e \) is given by \( F_e = qE \), where \( q \) is the charge of the alpha particle and \( E \) is the electric field between the plates.

Calculate the electric field \( E \) between the plates using the formula \( E = \frac{V}{d} \), where \( V \) is the potential difference across the plates (150 V) and \( d \) is the separation between the plates (8.20 mm).

The magnetic force \( F_m \) on a moving charge in a magnetic field is given by \( F_m = qvB \), where \( v \) is the velocity of the alpha particle and \( B \) is the magnetic field. For the particle to be undeflected, set \( F_e = F_m \), leading to \( qE = qvB \).

Solve for the velocity \( v \) of the alpha particles using the kinetic energy gained from the potential difference they were accelerated through: \( \frac{1}{2}mv^2 = qV_{acc} \), where \( V_{acc} \) is 1.75 kV. Rearrange to find \( v = \sqrt{\frac{2qV_{acc}}{m}} \).

Finally, solve for the magnetic field \( B \) using the equation \( B = \frac{E}{v} \). Substitute the expressions for \( E \) and \( v \) to find the magnitude of \( B \). The direction of \( B \) should be perpendicular to both the velocity of the particles and the electric field, following the right-hand rule.

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Electric Field

An electric field is a region around a charged object where other charged objects experience a force. In this scenario, the electric field is created by the parallel plates connected to a 150-V battery, influencing the motion of charged particles like alpha particles. The field's strength is determined by the voltage and separation between the plates.

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Percorso guidato

03:16

Intro to Electric Fields

Lorentz Force

The Lorentz force is the force experienced by a charged particle moving through electric and magnetic fields. It is the sum of the electric force and the magnetic force acting on the particle. For the alpha particles to remain undeflected, the magnetic force must counterbalance the electric force, requiring precise calculation of the magnetic field's magnitude and direction.

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Percorso guidato

13:39

Lorentz Transformations of Velocity

Magnetic Field

A magnetic field exerts a force on moving charged particles, influencing their trajectory. In this problem, the magnetic field must be oriented such that it cancels the deflection caused by the electric field on the alpha particles. The direction and magnitude of the magnetic field are crucial for ensuring the particles pass through the plates without deviation.

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Percorso guidato

05:30

Magnetic Fields and Magnetic Dipoles

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