Chapter 4: Q28E (page 913)

(a) What is the speed of a beam of electrons when the simultaneous influence of an electric field of 1.56 * 10⁴ V/m and a magnetic field of 4.62 * 10^-3 T, with both fields normal to the beam and to each other, produces no deflection of the electrons? (b) In a diagram, show the relative orientation of the vectors v, E, and B. (c) When the electric field is removed, what is the radius of the electron orbit? What is the period of the orbit?

Short Answer

Expert verified

The speed of the beam electrons is $3.88 \times 10^{6} m / s$
The relative orientation of vectors is

The radius of the orbit is $4.17 \times 10^{- 3}$ and period of the orbit is $7.74 \times 10^{- 9} s$

Step by step solution

Magnetic force

Magnetic force is given by

$\vec{F_{B}} = q (\vec{v} \times \vec{B})$

Determine the speed of beam of the electrons

(a)

Magnetic force is given by

$\vec{F_{B}} = q (\vec{v} \times \vec{B})$

Therefore, the speed of the beam is

$\begin{aligned} v = \frac{E}{B} \\ = \frac{1 .56 \times 10^{4} V / m}{4 .62 \times 10^{- 3} T} \\ = 3 .38 \times 10^{6} m / s \\ V = 3 .38 \times 10^{6} m / s \end{aligned}$

Determine the relative orientation of the vectors

The sketch of the vector’s velocity electric field and magnetic field is

Determine the radius of the orbit and period of orbit

There is 90-degree angle between the velocity and magnetic field

The force acting on the ion is

$\begin{aligned} F_{B} = | q | vBsin (ϕ) \\ = | q | vBsin (90^{\circ}) \\ = | q | vB \\ And the equation is \\ F = | q | vB = ma \\ | q | vB = m \frac{v^{2}}{R} \end{aligned}$

And the radius is

$\begin{aligned} R = \frac{mv}{| q | B} \\ = \frac{(9 .11 \times 10^{- 31} kg) (3 .38 \times 10^{6} m / s)}{(1 .60 \times 10^{- 10} C) (4 .62 \times 10^{- 3} T)} \\ = 4 .17 \times 10^{- 3} m \end{aligned}$

Therefore, the radius of the orbit is $4.17 \times 10^{- 3} m$

And the period of orbit is

$\begin{aligned} T = \frac{2 π}{ω} \\ = \frac{2 πR}{V} \\ = \frac{2 π (4 .17 \times 10^{- 3} m)}{(3 .38 \times 10^{6} m / s)} \end{aligned}$

Therefore, the period of the orbit is $7.74 \times 10^{- 9} s$

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Short Answer

Step by step solution

Magnetic force

Determine the speed of beam of the electrons

Determine the relative orientation of the vectors

Determine the radius of the orbit and period of orbit

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