Question

Figure 29-25 represents a snapshot of the velocity vectors of four electrons near a wire carrying current i. The four velocities have the same magnitude; velocity is directed into the page. Electrons 1 and 2 are at the same distance from the wire, as are electrons 3 and 4. Rank the electrons according to the magnitudes of the magnetic forces on them due to current i, greatest first.

Short answer

The ranking of electrons according to the magnitude of the magnetic forces on them due to current i is${\mathrm{F}}_1>{\mathrm{F}}_3={\mathrm{F}}_4>{\mathrm{F}}_2$.

Step-by-step solution

Step 1: Given

Figure representing a snapshot of the velocity vectors of four electrons near a wire carrying currenti is given.

$\left|{\mathrm{v}}_1\right|=\left|{\mathrm{v}}_2\right|=\left|{\mathrm{v}}_3\right|=\left|{\mathrm{v}}_4\right|=\mathrm{v}$

Determining the concept

Find the magnetic field on each electron using the formula for the magnetic field at a perpendicular distance from the wire. From this, find the magnetic forces acting on each electron. Comparing them, findthe ranking of electrons according to the magnitude of the magnetic forces on them due to current i.

Right Hand Rule 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 motion of the conductor relative to the magnetic field, and the forefinger points towards the direction of the magnetic field and the middle finger points towards the direction of the induced current.

Formulae are as follow:

$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi R}}$

role="math" localid="1663331285544" $\mathrm{F}=\mathrm{qvB}\sin \varphi$

Where, B is magnetic field, R is radius, iis current,𝛍is permeability,F is magnetic force, v is velocity, q is charge of particle.

Determining the rank of electrons according to the magnitude of the magnetic forces on them due to current i, greatest first

Magnetic field at a distance R from a wire carrying current i is given by,

$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi R}}$

Let electron 1 and 2 be at distance r and electron 3 and 4 be at distance R from the wire. Then, the magnetic fields experienced by the given electrons are,

${\mathrm{B}}_1=\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi r}},{\mathrm{B}}_2=\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi r}},{\mathrm{B}}_3=\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi R}},{\mathrm{B}}_4=\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi R}}$

Since, R > r,

${\mathrm{B}}_3={\mathrm{B}}_4<{\mathrm{B}}_1={\mathrm{B}}_2$

From right hand rule, interpret that magnetic field is directing into the page.

The magnetic force on electron is given by,

$\mathrm{F}=\mathrm{evB}\mathrm{sin\varphi }$

From right hand rule, interpret that magnetic field on each electron is directed into the page.

Hence, the magnetic forces on the given electrons are,

${\mathrm{F}}_1={\mathrm{evB}}_1\sin 90°=\mathrm{ev}\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi r}}$, ${\mathrm{F}}_2={\mathrm{evB}}_2\sin 180°=0$, ${\mathrm{F}}_3={\mathrm{evB}}_3\sin 90°=\mathrm{ev}\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi R}}$

${\mathrm{F}}_4={\mathrm{evB}}_4\sin 90°=\mathrm{ev}\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi R}}$

Hence, the ranking of electrons according to the magnitude of the magnetic forces on them due to currenti is,

${\mathrm{F}}_1>{\mathrm{F}}_3={\mathrm{F}}_4>{\mathrm{F}}_2$

Therefore, magnetic force on the particle due to currenti in the wire depends on the magnetic field, its distance from the wire, and the angle between its velocity and applied magnetic field.