If$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}\mathbf{=}\mathit{q}\left(\stackrel{\rightharpoonup }{v}x\stackrel{\rightharpoonup }{B}\right)\mathbf{}\mathbf{and}\mathbf{}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}$is perpendicular to$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{B}}$, then what is the direction of B in the three situations shown in Fig. 3-24 when constant q is (a) positive and (b) negative?

$$\begin{gathered} \stackrel{\rightharpoonup }{F}=q\left(\stackrel{\rightharpoonup }{v}x\stackrel{\rightharpoonup }{B}\right) \\ \stackrel{\rightharpoonup }{v}\perp \stackrel{\rightharpoonup }{B} \end{gathered}$$

The problem is based on the right-hand rule. It determines the directions of magnetic force, conventional current, and magnetic field. It states that, if we arrange our thumb, forefinger, and middle finger of the right-hand perpendicular to each other, then the thumb points towards the direction of the magnetic force, the forefinger points towards the direction of the magnetic field, and the middle finger points towards the direction of the conventional current. Here formula of the expression of the vector product of two vectors can be used along with the right-hand rule.

Formula:

$$\begin{gathered} \stackrel{\rightharpoonup }{v}x\stackrel{\rightharpoonup }{B}=vB\sin \sin \theta \\ \stackrel{\rightharpoonup }{F}=q\left(\stackrel{\rightharpoonup }{v}x\stackrel{\rightharpoonup }{B}\right) \end{gathered}$$

We have given,$\stackrel{\rightharpoonup }{v}\perp \stackrel{\rightharpoonup }{B}$the maximum magnetic force acting on the charge q. For it$$\begin{gathered} \theta =90° \\ \mathrm{Hence},\stackrel{\rightharpoonup }{\mathrm{v}}\mathrm{x}\stackrel{\rightharpoonup }{\mathrm{B}}=\mathrm{vB} \end{gathered}$$.

Hence,

a) For positive charge:

The right-hand rule gives the direction of$\left(\stackrel{\rightharpoonup }{v}x\stackrel{\rightharpoonup }{B}\right)$as the direction of the thumb. If q is positive, then the direction of$\stackrel{\rightharpoonup }{F}=q\left(\stackrel{\rightharpoonup }{v}x\stackrel{\rightharpoonup }{B}\right)$is in the direction of$\left(\stackrel{\rightharpoonup }{v}x\stackrel{\rightharpoonup }{B}\right)$.

Figure (1)$\stackrel{\rightharpoonup }{v}$is acting along the +z axis and$\stackrel{\rightharpoonup }{F}$is acting along the positive y-axis. Here$\stackrel{\rightharpoonup }{v}$can be swept in the counterclockwise direction in such a way that$\stackrel{\rightharpoonup }{F}$is going upward direction.

Hence,$\stackrel{\rightharpoonup }{B}$is acting along the positive x-axis.

Figure (2) $\stackrel{\rightharpoonup }{v}$is acting along the negative x-axis and $\stackrel{\rightharpoonup }{F}$is acting along the positive y-axis. Here can be swept in the counterclockwise direction in such a way that $\stackrel{\rightharpoonup }{F}$is going upward direction.

Hence, $\stackrel{\rightharpoonup }{B}$is acting along the positive z-axis.

Figure (3)$\stackrel{\rightharpoonup }{v}$is acting along the positive x-axis and$\stackrel{\rightharpoonup }{F}$ is acting along the negative y-axis. Hence$\stackrel{\rightharpoonup }{v}$can be swept in the clockwise direction in such a way that$\stackrel{\rightharpoonup }{F}$is going in the downward direction.

Hence, $\stackrel{\rightharpoonup }{B}$is acting along the positive z-axis

b) For negative charge:

If q is negative,$\stackrel{\rightharpoonup }{F}=q\left(\stackrel{\rightharpoonup }{v}x\stackrel{\rightharpoonup }{B}\right)$ then the direction ofis in the opposite direction of$\left(\stackrel{\rightharpoonup }{v}x\stackrel{\rightharpoonup }{B}\right)$.

So the direction of$\stackrel{\rightharpoonup }{B}$should be opposite to all those answers in part a).

So figure (1),$\stackrel{\rightharpoonup }{B}$is acting along the negative x-axis.

Figure (2),$\stackrel{\rightharpoonup }{B}$is acting along the negative z-axis.

Figure (3), $\stackrel{\rightharpoonup }{B}$is acting along the negative z-axis.