Force Acting On a Current Carrying Conductor In a Magnetic Field, Force Acting Between Two Conductor Carrying Conductor (JEE LEVEL)

Let us consider two isolated current elements {{i}_{1}}d{{l}_{1}} and {{i}_{2}}d{{l}_{2}}. The magnetic field due to 1 at the site of 2 is given by Biot-Savart law is

                        d{{\vec{B}}_{{12}}}=\frac{{{{\mu }_{o}}}}{{4\pi }}\frac{{{{i}_{1}}d{{{\vec{l}}}_{1}}\times {{{\vec{r}}}_{{12}}}}}{{r_{{12}}^{3}}}

         The force experienced by a current element in a magnetic field is given by Lorentz force d\vec{F}=id\vec{l}\times \vec{B}

               Force experienced by 2 due to 1={{i}_{2}}d{{\vec{l}}_{2}}\times \frac{{{{\mu }_{o}}}}{{4\pi }}\frac{{{{i}_{1}}d{{{\vec{l}}}_{1}}\times {{{\vec{r}}}_{{12}}}}}{{r_{{12}}^{3}}}

                               d{{\vec{F}}_{{12}}}=\frac{{{{\mu }_{o}}}}{{4\pi }}{{i}_{1}}{{i}_{2}}\frac{{d{{{\vec{l}}}_{2}}\times (d{{{\vec{l}}}_{1}}\times {{{\vec{r}}}_{{12}}})}}{{r_{{12}}^{3}}}

         Similarly             d{{\vec{F}}_{{21}}}=\frac{{{{\mu }_{o}}}}{{4\pi }}{{i}_{1}}{{i}_{2}}\frac{{d{{{\vec{l}}}_{1}}\times (d{{{\vec{l}}}_{2}}\times {{{\vec{r}}}_{{21}}})}}{{r_{{21}}^{3}}}

 In magnitude the forces can be written as

                                  |d{{\vec{F}}_{{12}}}|\,\,\,=\frac{{{{\mu }_{o}}}}{{4\pi }}{{i}_{1}}{{i}_{2}}\frac{{|d{{{\vec{l}}}_{1}}|\,\,\,|d{{{\vec{l}}}_{2}}|}}{{r_{{21}}^{2}}}\cos {{\theta }_{2}}

and                               

                                    |d{{\vec{F}}_{{21}}}|\,\,\,=\frac{{{{\mu }_{o}}}}{{4\pi }}{{i}_{1}}{{i}_{2}}\frac{{|d{{{\vec{l}}}_{1}}|\,\,\,|d{{{\vec{l}}}_{2}}|}}{{r_{{21}}^{2}}}\cos {{\theta }_{1}}

\displaystyle since\text{ }{{\theta }_{1}}\ne {{\theta }_{2}}

\displaystyle hence\text{ }|d{{\vec{F}}_{{12}}}|\ne d{{\vec{F}}_{{21}}}|

         Thus we see that the forces are not equal and opposite. Hence there          is apparent violation of the third law of motion. This does not mean that there is breakdown of Newton’s third law.

         This apparent violation is due to acceleration and retardation of charge carriers at the ends of the elements. In the isolated elements charge carries are accelerated at the starting end from zero to a uniform velocity and at the other end from that uniform velocity to zero.

         This acceleration and retardation produce and electromagnetic field, which carries away momentum from charge carriers, that is, charge carriers experience additional force. This force compounded with the above force shows that the force exerted by the elements are equal and opposite. So violation is only apparent. This situation arises only in case of isolated current element where there are acceleration and retardation of the charge carriers at the ends. In case of loops where there is no such acceleration or retardation of charge carriers there should be no violation of third law. In fact such difficulty does not arise. For loops shown in the figure here.

            \displaystyle {{\vec{F}}_{{12}}}=\frac{{{{\mu }_{o}}}}{{4\pi }}{{i}_{1}}{{i}_{2}}\underset{{loop1}}{\mathop{\oint }}\,\text{ }\underset{{loop2}}{\mathop{\oint }}\,\frac{{d{{{\vec{l}}}_{2}}\times (d{{{\vec{l}}}_{1}}\times {{{\vec{r}}}_{{12}}})}}{{r_{{12}}^{2}}}

         By mathematical laws this can be transformed into symmetrical from      

              \displaystyle {{\vec{F}}_{{12}}}=\frac{{{{\mu }_{o}}}}{{4\pi }}{{i}_{1}}{{i}_{2}}\underset{{Loop1}}{\mathop{\oint }}\,\text{ }\underset{{Loop2}}{\mathop{\oint }}\,\frac{{(d{{{\vec{l}}}_{1}}.d{{{\vec{l}}}_{2}}){{{\vec{r}}}_{{12}}}}}{{r_{{12}}^{2}}}

         This equation shows that {{\vec{F}}_{{12}}}=-{{\vec{F}}_{{21}}}.