Biot Savart Law is given as : \begin{equation} \mathbf{B}\left(\mathbf{r}\right)=\oint\limits_{\textrm{around}}^{\textrm{wire}}\dfrac{I\,\mathrm{d}\boldsymbol{\ell'}\boldsymbol{\times}\left(\mathbf{r}-\mathbf{r'}\right)}{\vert\mathbf{r}-\mathbf{r'} \vert^{3}} \tag{01} \end{equation} The book that I'm following says that if we take \begin{equation} \mathrm{d}\mathbf{B}\left(\mathbf{r}\right)=\dfrac{I\,\mathrm{d}\boldsymbol{\ell'}\boldsymbol{\times}\left(\mathbf{r}-\mathbf{r'}\right)}{\vert\mathbf{r}-\mathbf{r'} \vert^{3}} \tag{02} \end{equation} for a current element $\,I\mathrm{d}\boldsymbol{\ell'}\,$, then this will not be a correct assumption since the force exerted by $\,\mathrm{d}\mathbf{B}\,$ on a charge $\,q\,$ at position $\,\mathbf{r}\,$ moving with velocity $\,\mathbf{V}\,$ is
\begin{equation} \mathrm{d}\mathbf{F}=\dfrac{Iq\mathbf{V}}{c}\boldsymbol{\times}\dfrac{\mathrm{d}\boldsymbol{\ell'}\boldsymbol{\times}\left(\mathbf{r}-\mathbf{r'}\right)}{\vert\mathbf{r}-\mathbf{r'} \vert^{3}} \tag{03} \end{equation}
and since "this force violates Newton's Third Law. It is not directed along the line joining the current element with the charge." I don't get this quoted line. How is Newton's Third law acting here?
I can give you the exact texts of the book.Here
Here is a general outline of the solution which I thought for the question
1.See that Newton's third law is connected to the conservation of momentum principle.
2.Now evaluate the $\mathbf E$ and $\mathbf B$ fields and hence $\mathbf F_B$ and $\mathbf F_E$ by the corresponding laws.
3.See that net force is not conserved which implies momentum should not be conserved.
4. Hence we cannot use the expression of '$d \mathbf B$' for the line element
