Chapter 4: Q25E (page 913)

A proton (q = 1.60 * 10^-19 C, m = 1.67 * 10^-27 kg) moves in a uniform magnetic field. At t = 0 the proton has velocity components v_x = 1.50 * 10⁵ m/s, v_y = 0, and v_z = 2.00 * 10⁵ m/s (see Example 27.4). (a) What are the magnitude and direction of the magnetic force acting on the proton? In addition to the magnetic field there is a uniform electric field in the +x-direction $\vec{E} = (+ 2.00 x 10 \frac{V}{m}) \hat{i}$ . (b) Will the proton have a component of acceleration in the direction of the electric field? (c) Describe the path of the proton. Does the electric field affect the radius of the helix? Explain. (d) At t = T/2, where T is the period of the circular motion of the proton, what is the x-component of the displacement of the proton from its position at t = 0?

Short Answer

Expert verified

The force acting on the proton is $(1 .60 \times 10^{- 14} N) \hat{j}$
Yes, the proton has a component of acceleration in the direction of the electric field.
The path of the proton is helix.
The x-component of the displacement of the proton from its position at t = 0 is 1.40cm

Step by step solution

Force acting on proton

The total force acting on a proton is given by

$\vec{F} = q (\vec{E} + \vec{v} \times \vec{B}) = q (\vec{v} \times \vec{B})$

Determine the force acting on a proton

(a)

The total force acting on a proton is

$\begin{aligned} \vec{F} = q (\vec{E} + \vec{v} \times \vec{B}) = q (\vec{v} \times \vec{B}) \\ = q ([v_{x} \hat{i} + v_{z} \hat{k}] \times [B_{x} \hat{i}]) \\ = {qv}_{z} B_{x} \hat{j} \\ = (1 .60 \times 10^{- 19} C) (2 .00 \times 10^{5} m / s) (0 .500 T) \hat{j} \\ = (1 .60 \times 10^{- 14} N) \hat{j} \end{aligned}$

Therefore, the force acting on the proton is $(1 .60 \times 10^{- 14} N) \hat{j}$

(b)

The electric field is $\vec{E} = (+ 2 .00 \times 10^{4} V / m) \hat{i}$

Therefore, the force due to the electric field will be at same direction,

So, yes, the proton has a component of acceleration in the direction of the electric field

Determine the path of the proton

(c)

The motion is circular due to the magnetic field and the magnetic field is perpendicular to the electric field,

Therefore, the motion will be circular but in x direction.

So, the path of the proton is helix.

Determine the x-component displacement of proton

(d)

The period of motion is

$\begin{aligned} T = \frac{2 π}{ω} \\ = \frac{2 πm}{| q | B} \\ Therefore, \\ T = \frac{2 π (1 .67 \times 10^{- 27} kg)}{(1 .60 \times 10^{- 19} C) (0 .500 T)} \\ = 1 .312 \times 10^{- 7} s \\ And we know that acceleration is \\ a_{x} = \frac{eE}{m} \end{aligned}$

And the formula of the displacement will be

$x - x_{0} = v_{0 x} t + \frac{1}{2} a_{x} t^{2}$

Put the values in the formula

$\begin{aligned} x - x_{0} = v_{0 x} t + \frac{1}{2} a_{x} t^{2} \\ = \frac{1}{2} v_{0 x} T + \frac{1}{8} (\frac{F_{E}}{m}) T^{2} \\ = \frac{1}{2} (1 .50 \times 10^{5} m / s) (1 .312 \times 10^{- 7} s) + \frac{1}{8} (\frac{(1 .60 \times 10^{- 19} C) (2 .00 \times 10^{4} V / m)}{1 .67 \times 10^{- 27} kg}) {(1 .312 \times 10^{- 7} s)}^{2} \\ = 1 .40 \times 10^{- 2} m = 1 .40 cm \end{aligned}$