Chapter 4: Q29E (page 913)

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 * 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?

Short Answer

Expert verified

The magnitude 0.0445T and the direction out of the pageis needed to that alpha particle emerge undeflected from between the plates.

Step by step solution

Potential energy and kinetic energy of a charged particle in an electric field

The potential energy and the kinetic energy are equal

Therefore,

$\frac{1}{2} {mv}^{2} = qV$

And

$v = \sqrt{\frac{2 qV}{m}}$

Determine the magnitude and direction of the particle

The potential energy and the kinetic energy are equal

$\begin{aligned} \frac{1}{2} {mv}^{2} = qV \\ v = \sqrt{\frac{2 qV}{m}} \end{aligned}$

Therefore,

$\begin{aligned} v = \sqrt{\frac{2 qV}{m}} \\ = \sqrt{\frac{4 (1 .60 \times 10^{- 19} C) (1750 V)}{6 .64 \times 10^{- 27} kg}} \\ = 4 .11 \times 10^{5} m / s \end{aligned}$

And the magnetic force is equal to the electric force

Therefore,

$\begin{aligned} qvB = qE \\ B = \frac{E}{v} \end{aligned}$

And the electric field is

$\begin{aligned} E = \frac{E}{d} \\ = \frac{150 V}{0 .00820 m} \\ = 18300 V / m \end{aligned}$

And the magnetic field is

$\begin{aligned} B = \frac{E}{V} \\ = \frac{18300 V / m}{4 .11 \times 10^{5} m / s} \\ = 0 .0445 T \end{aligned}$

Therefore, And the direction of the magnetic field is out of the page and the magnitude is 0.0445T

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Short Answer

Step by step solution

Potential energy and kinetic energy of a charged particle in an electric field

Determine the magnitude and direction of the particle

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