A wire lying along a yaxis from y=0to y=0.250mcarries a current of 2.00mAin the negative direction of the axis. The wire fully lies in a nonuniform magnetic field that is given by$\mathbf{B}\mathbf{}\mathbf{⃗}\mathbf{=}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{}\mathbf{T}\mathbf{/}\mathbf{m}\mathbf{}\mathbf{)}\mathbf{}\mathbf{y}\hat{\mathbf{i}}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{4}\mathbf{}\mathbf{T}\mathbf{/}\mathbf{m}\mathbf{}\mathbf{)}\mathbf{y}\hat{\mathbf{j}}$

In unit-vector notation, what is the magnetic force on the wire?

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

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 $d\overrightarrow{F}$and integrating it with respective to y, we can find themagnetic force on the wire.

Formula:

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

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

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

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

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

$dF⃗=idl(-\hat{j})\times (0.3yî+0.4y\hat{j})$

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

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

Let,$\xi =0.6\times {10}^{-3}N/m^2$

Therefore,

$dF⃗=\xi ydl\hat{k}$

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

$$\begin{gathered} \overrightarrow{F}=\int d\overrightarrow{F} \\ =\xi \hat{k}{\int }_0^{0.25}ydy \\ \overrightarrow{F}=\xi \hat{k}\left(\frac{{\left(0.25m\right)}^2}{2}\right) \end{gathered}$$

Therefore,

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