Chapter 4: Q51E (page 915)

When a particle of charge q > 0 moves with a velocity ofat 45.0° from the +x-axis in the xy-plane, a uniform magnetic field exerts a force along the -z-axis (Fig. P27.51). When the same particle moves with a velocitywith the same magnitude asbut along the +z-axis, a forceof magnitude F₂ is exerted on it along the +x-axis. (a) What are the magnitude (in terms of q, v₁, and F₂) and direction of the magnetic field? (b) What is the magnitude ofin terms of F₂?

Short Answer

Expert verified

The magnitude and direction of magnetic field is ${\vec{B}}_{y} = - \frac{F_{2}}{v_{1} q} \hat{y}$
The magnitude of is ${\vec{F}}_{1} is F_{1} = \frac{F_{2}}{\sqrt{2}}$

Step by step solution

Magnetic force and electric force

The magnetic force is given by

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

The electric force is given by

$\vec{F_{E}} = q \vec{E}$

Determine the magnetic field

(a)

The magnetic force is given by

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

Now the magnetic force is

$\begin{aligned} {\vec{F}}_{2} = F_{2} \hat{x} \\ {\vec{v}}_{2} = v_{1} \hat{k} \\ F_{2} \hat{x} = {qv}_{1} \hat{k} \times \vec{B} \\ = - {qB}_{y} v_{1} \hat{x} + {qB}_{x} v_{1} \hat{y} + 0 \hat{z} \\ B_{x} = 0 \\ F_{2} = - {qv}_{1} B_{y} \end{aligned}$

Therefore, the magnetic field is ${\vec{B}}_{y} = - \frac{F_{2}}{V_{1} q} \hat{y}$

Determine the magnitude of F1→

(b)

The magnitude of $\vec{F_{1}}$ is

role="math" localid="1668264415915" $\begin{aligned} {\vec{F}}_{1} = - F_{1} \hat{z} \\ {\vec{v}}_{1} = v_{1} \cos (45) \hat{x} + v_{1} \sin (45) \hat{y} \\ - F_{1} \hat{z} = q {\vec{v}}_{1} \times \vec{B} \\ = {qv}_{1} \sin (45) B_{z} \hat{x} - {qv}_{1} \cos (45) B_{z} \hat{y} + ({qv}_{1} \cos (45) B_{y} - {qv}_{1} \sin (45) B_{x}) \hat{z} \\ Now, \\ B_{z} = 0 \\ B_{x} = 0 \\ - F_{1} = B_{y} {qv}_{1} \cos (45) \\ F_{1} = - B_{y} {qv}_{1} \cos (45) \\ = - (- \frac{F_{2}}{v_{1} q}) {qv}_{1} \cos (45) \\ F_{1} = F_{2} \cos (45) = \frac{F_{2}}{\sqrt{2}} \end{aligned}$