91Ó°ÊÓ

Current is in the right direction, and four different choices of the magnetic field are given.

Consider the formula for the force:

$\stackrel{→}{{\mathbf{F}}_B}\mathbf{=}\mathbf{i}(\stackrel{\mathbf{→}}{\mathbf{l}}\mathbf{×}\stackrel{\mathbf{→}}{\mathbf{B}})\mathbf{W}$

Here, F_B is magnetic force, i is current, l is length, B is magnetic field.

Ranking of the given choices:

For choice 1, the magnetic field is along the same direction as the current flowing in the wire and between the current elements, the magnetic field is zero.

Hence, no force will act on the charges and the resultant force will be zero. To get the force, the non-zero current element and magnetic field must be perpendicular to each other.

$$\begin{gathered} \stackrel{→}{F_B}=i(\stackrel{→}{l}×\stackrel{→}{B})=ilB\sin \mathrm{θ} \\ \stackrel{→}{F_B}=ilB\sin 0 \\ \stackrel{→}{F_B}=0 \end{gathered}$$

Then,

$E=0$

Hence,

$V=Ed=\left(0\right)d=0$

For choice 2, the same thing happens as choice 1 but the direction of magnetic field and current is opposite and the angle between them is$180°$and$\sin 180=0$.So, in this case

$\stackrel{→}{F_B}=0$

Then,

$E=0$

Hence,

$V=Ed=\left(0\right)d=0$

For choice 3 the magnetic field is inside the plane. As most of the charge carriers in the wire are electrons and using the right hand rule, the force will act on electrons in the upward direction and electrons will accumulate on the upper side of the wire leaving behind the positive charges. Hence, the upper side of the wire will be at a lower potential and lower side will be at high potential.

$$\begin{gathered} {\stackrel{→}{F}}_B=q(\stackrel{→}{V}×\stackrel{→}{B})=qvB\sin \mathrm{θ} \\ \mathrm{θ}=90° \\ {\stackrel{→}{F}}_B=qvB\sin 90=qVB \\ E=\frac{F_B}{q}=VB \\ V=Ed=vBd \end{gathered}$$

For choice 4:

The magnetic field is outside the plane. As most of the charge carriers in the wire are electrons and using the right hand rule, the force will act on electrons in the downward direction and electrons will accumulate on lower side of the wire leaving behind the positive charges. Hence, upper side of the wire will be at higher potential and lower side will be at lower potential.

$$\begin{gathered} {\stackrel{→}{F}}_B=q(\stackrel{→}{V}×\stackrel{→}{B})=qvB\sin \mathrm{θ} \\ \mathrm{θ}=90° \\ {\stackrel{→}{F}}_B=qvB\sin 90=qVB \\ E=\frac{F_B}{q}=VB \\ V=Ed=vBd \end{gathered}$$

Hence, choices 3 and 4 have the same value of potential difference, and then the choice 1 and 2 have the same value of potential difference.

From part a) it can be concluded that, for choice 4, the top side of the wire will be at higher potential.

Hence, for choice 4, the top side of the wire will be at higher potential.

Therefore, using the right hand rule rank the choices for magnetic field according to the magnitude of the electric potential differences.