Question

Figure 28-24 shows a metallic, rectangular solid that is to move at a certain speed vthrough the uniform magnetic field$\overrightarrow{\mathbf{B}}$. The dimensions of the solid are multiples of d, as shown.You have six choices for the direction of the velocity: parallel to x, y, or zin either the positive or negative direction.

(a) Rank the six choices according to the potential difference set up across the solid, greatest first.

(b) For which choiceis the front face at lower potential?

Short answer

a.Ranking according to the magnitude of the voltage across the solids:

$V_x=V_{-x}<V_y=V_{-y}<V_z=V_{-z}$

b. For velocity in $+y$direction, the front face will be at a lower potential.

Step-by-step solution

Step 1: Given

Magnetic field B is along x direction.

Dimensions of rectangular boxes are given in the figure.

Determining the concept

Use the formula of magnetic force to determine the force on the charges with different directions of velocities. Then from that, find the electric field. Then using the relation between the electric field and voltage, find the voltage across the solids corresponding to different direction’s velocities.

Formulae are as follow:

The force acting on the charge is-

$\begin{array}{c}\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})\\ =qvBsin\theta .\widehat{n}\\ E=\frac{F_B}{q}\\ V=Ed\end{array}$

Where, F is magnetic force, v is velocity, E is electric field, B is magnetic field, q is charge of particle, d is distance.

(a) Determining the six choices of velocity according to the potential difference developed across the solid.

Inthegiven figure, uniform magnetic field$\left(B\right)$is along positive x direction, and velocity of rectangular solid$\left(v\right)$is also along$+x$direction. The angle between these two vectors is zero, and the force acting on the charge will be zero. Hence potential difference could not be developed for this choice of velocity.

$\begin{array}{c}{F}_B=q(v\times B)\\ =qvB\sin \theta \\ =qvB\sin 0°\\ =0\end{array}$

$\begin{array}{c}E=\frac{F_B}{q}\\ =0\end{array}$

$\begin{array}{c}{V}_x=Ed\\ =0\end{array}$

Similarly, velocity of rectangular solid$\left(v\right)$along $-x$direction will cause no force on charges, and the force acting on the charges is zero.

Here, the angle between magnetic field $\left(B\right)$and velocity$\left(V\right)$is.$180°$

$V_{-x}=0$

When the velocity is along $+y$direction and magnetic field $\left(B\right)$along$+x$direction, the force will act on the charge particles because velocity and magnetic field are perpendicular to each other, and a non-zero force will be given.

$\begin{array}{c}{\overrightarrow{F}}_B=-e(v\widehat{y}\times B\widehat{x})\\ =evB\widehat{z}\\ \{\because (\sin 90°)=1\}\\ \begin{array}{c}E=\frac{F_B}{q}\\ =\frac{evB}{e}\\ =vB\end{array}\end{array}$

$\begin{array}{c}{V}_y=Ed\\ =dvB\end{array}$

The force will act on negative charges along$+z$direction. Hence the back face will beat positive potential, and front face will be at negative potential.

And when velocity is along the$-y$direction, the force on the electrons will act alongdirection (inward to the plane). Thenthefront face will be at positive potential, and the back face will be at negative potential.

$$\begin{gathered} F_B=qvB\sin 90°=qvB \\ E=\frac{F_B}{q}=vB \end{gathered}$$

The potential difference ($v$) is then,

$\begin{array}{c}{V}_{-y}=Ed\\ =vBd\end{array}$

And when the velocity is along positive z direction and magnetic field B along x direction, the force on the electrons will act along negative y direction. Hence the lower face will be at negative potential, and upper face will be at higher potential

$$\begin{gathered} \overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B}) \\ \begin{array}{c}\overrightarrow{F}=-e(v\widehat{Z}\times B\widehat{x})\\ =evB(-\widehat{y})\end{array} \\ \begin{array}{c}E=\frac{F_B}{q}\\ =\frac{evB}{e}\\ =vB\end{array} \end{gathered}$$

$\begin{array}{c}{V}_Z=Ed\\ =vB\left(2d\right)\\ =2dvB\end{array}$

Similarly, when velocity is along$-z$direction,

$\begin{array}{c}\overrightarrow{F}=-e(v(-\widehat{Z})\times B\widehat{x})\\ =evB\left(\widehat{y}\right)\end{array}$

The electrons will deflect totheupper face. Hence the upper face will be at negative potential, leaving the lower face at positive potential.

$\begin{array}{c}E=\frac{F}{q}\\ =\frac{evB}{e}\\ =vB\end{array}$

$\begin{array}{c}{V}_{-z}=Ed\\ =vB\left(2d\right)\\ =2dvB\end{array}$

Therefore,ranks according to the magnitude of the voltage across the solids:

$V_x=V_{-x}<V_y=V_{-y}<V_z=V_{-z}$

(b) Determining the for which choice will the front face be at a lower potential

From part a), it can be concluded that when the velocity of rectangular box will be at $+y$direction, the front face will be at a lower potential.