91Ó°ÊÓ

1. The wire is lying along y axis from y=0 to y=0.250m.
2. The current $i=2.00mAor2.00×{10}^{-3}A$

3. The uniform magnetic field $B⃗=(0.3T/m)y\hat{i}+(0.4T/m)y\hat{j}.$

The influence of a magnetic field produced by one charge on another is what is known as the magnetic force between two moving charges. By using the differential equation for force $d\stackrel{→}{F}$and integrating it with respective to y, we can find themagnetic force on the wire.

Formula:

Magnetic force acting on current element $\stackrel{→}{idL}$is

$d\left(F_B\right)⃗=i\left(dl\right)⃗×B⃗$

Magnetic force acting on current element $\stackrel{→}{idL}$is given by

$d\left(F_B\right)⃗=i\left(dl\right)⃗×B⃗$

Since the unit vector associated with the current element is$-\hat{j}$

Also,$B⃗=(0.3T/m)y\hat{i}+(0.4T/m)y\hat{j}$

Therefore, the force on the current element is

$dF⃗=idl(-\hat{j})×(0.3yi\mathrm{̂}+0.4y\hat{j})$

Since$\hat{j}×\hat{i}=-\hat{k}$and $\hat{j}×\hat{j}=0$, we obtain

localid="1662964246711" $\begin{array}{l}d\stackrel{→}{F}=0.3iydI\hat{k}\\ =0.3×(2×{10}^{-3}A)+ydl\hat{k}\\ =(0.6×{10}^{-3}N/m^2)×ydl\hat{k}\end{array}$$\begin{array}{l}d\stackrel{→}{F}=0.3iydlk\\ =0.3×(2×{10}^{-3}A)+ydl\hat{k}\\ =(0.6×{10}^{-3}{\mathrm{Nm}}^2)×ydl\hat{k}\end{array}$

Let,$\mathrm{ξ}=0.6×{10}^{-3}N/m^2$

Therefore,

$dF⃗=\mathrm{ξ}ydl\hat{k}$

Integrating with respect to y within the limits y=0 to y=0.250m, we get

$$\begin{gathered} \stackrel{→}{F}=∫d\stackrel{→}{F} \\ =\mathrm{ξ}\hat{k}{∫}_0^{0.25}ydy \\ \stackrel{→}{F}=\mathrm{ξ}\hat{k}\left(\frac{{\left(0.25m\right)}^2}{2}\right) \end{gathered}$$

Therefore,

$$\begin{gathered} \stackrel{→}{F}=(0.6×{10}^{-3}N/m^2)×\left(\frac{{\left(0.25m\right)}^2}{2}\right)\hat{k} \\ =\left(1.88×{10}^{-5}N\right)\hat{k} \end{gathered}$$