Question: A wire lying along an xaxis from$\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$to $\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\text{m}}$carries a current of $\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{\text{A}}$ in the positive xdirection. The wire is immersed in a nonuniform magnetic field that is given by localid="1663064054178" $\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{00}\frac{\mathbf{T}}{{\mathbf{m}}^{\mathbf{2}}}\mathbf{)}{\mathbf{x}}^{\mathbf{2}}\hat{\mathbf{i}}\mathbf{-}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{600}\frac{\mathbf{T}}{{\mathbf{m}}^{\mathbf{2}}}\mathbf{)}{\mathbf{x}}^{\mathbf{2}}\hat{\mathbf{j}}$ In unit-vector notation, what is the magnetic force on the wire?

$i=3A$

$L=1 \text{cm} \text{or} 0.01 \text{m}$

$\overrightarrow{\mathrm{B}}=(4.00\frac{\mathrm{T}}{{\mathrm{m}}^2}){\mathrm{x}}^2\hat{\mathrm{i}}-(0.600\frac{\mathrm{T}}{{\mathrm{m}}^2}){\mathrm{x}}^2\hat{\mathrm{j}}$

$\overrightarrow{E}=300\hat{j}V/m$

$q=5\times {10}^{-6}\text{C}$

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 relation between magnetic force and magnetic field, we can find the magnetic force on the wire. We can find the force acting on the current element in the magnetic field.

$\overrightarrow{F}=i\overrightarrow{L}\times \overrightarrow{B}$

A straight wire carrying a current i in a uniform magnetic field experiences a sideways force as

$\overrightarrow{F}=i\overrightarrow{L}\times \overrightarrow{B}$

The force acting on the current element $id\overrightarrow{L}$in a magnetic field is

$d\overrightarrow{F}=id\overrightarrow{L}\times \overrightarrow{B}$

So calculate the total force by integrating the given equation

$\overrightarrow{F}=i\int \overrightarrow{B}dL$

By substituting the value, we can get

$$\begin{gathered} \overrightarrow{F}=i{\int }_0^1\hat{i}\times \left[\left(4.00\frac{T}{m^2}\right)x^2\hat{i}-\left(0.600\frac{T}{m^2}\right)x^2\hat{j}\right]dx \\ =3 A\times {\int }_0^1\left[-\left(0.600\frac{T}{m^2}\right)x^2\right]\left(\hat{i}\times \hat{j}\right)dx \\ =3 A\times {\int }_0^1\left[-\left(0.600\frac{T}{m^2}\right)x^2\right]\left(\hat{i}\times \hat{j}\right)dx \\ =3 A\times \left[-\frac{\left(0.600\frac{T}{m^2}\right)x^3}{3}\right]\left(\hat{k}\right) \\ =3 A\times {{\left[-\frac{\left(0.600\frac{T}{m^2}\right)x^3}{3}\right]}_0}^1\left(\hat{k}\right) \\ =\overrightarrow{F}=-\left(0.600N\right)\hat{k} \end{gathered}$$