The force on element $ dl $ in the figure at a distance $ l $ from a long wire carrying current $ {I_1} $ is

(A) $ \dfrac{{{\mu _0}{I_1}{I_2}}}{{2\pi }}\log \left( {dl} \right) $
(B) $ \dfrac{{{\mu _0}{I_1}{I_2}dl}}{{2\pi l}} $
(C) $ \dfrac{{{\mu _0}{I_1}{I_2}}}{{4\pi }}\log \left( {dl} \right) $
(D) $ \dfrac{{{\mu _0}{I_1}{I_2}}}{{4\pi l}}\log \left( {dl} \right) $

Hint: To solve this question, we need to evaluate the magnetic field at the point where the element is placed. Then, applying the formula of the force on a current carrying conductor, we will get the answer.

Formula used: The formula which is used in solving this question is given by
 $ B = \dfrac{{{\mu _0}I}}{{2\pi r}} $ , here $ B $ is the magnetic field produced by a straight current carrying conductor at a perpendicular distance of $ r $ from it.
 $ F = I\left( {\vec l \times \vec B} \right) $ , here $ F $ is the force acting on a conductor of length $ l $ carrying a current $ I $ , which is placed in a magnetic field of $ B $ .

Complete step by step solution:
We know that the magnetic field due to a straight current carrying wire is given by
 $ B = \dfrac{{{\mu _0}I}}{{2\pi r}} $ (1)
According to the question, we have $ I = {I_1} $ . Also, the element $ dl $ is at a distance $ r = l $ from the current carrying wire. So putting these in (1) we get
 $ B = \dfrac{{{\mu _0}{I_1}}}{{2\pi l}} $
From the right hand rule, we get the direction of this magnetic field into the plane of paper.
We know that the force on a current carrying conductor placed in a magnetic field is given by $ F = I\left( {\vec l \times \vec B} \right) $
Now, the current in the element is equal to $ {I_2} $ . Also, the length of this element is $ dl $ . So the force on this element is
 $ F = {I_2}\left( {d\vec l \times \vec B} \right) $
As the magnetic field is perpendicular to the current in the element, so we get
 $ F = {I_2}Bdl $
From (1)
 $ F = {I_2}\left( {\dfrac{{{\mu _0}{I_1}}}{{2\pi l}}} \right)dl $
 $ F = \dfrac{{{\mu _0}{I_1}{I_2}dl}}{{2\pi l}} $
Hence, the correct answer is option B.

Note:
We should not get confused as to why we have taken the magnetic field as uniform for the element $ dl $ . As it is of infinitesimally small length, so the variation of the magnetic field over its length is negligible. So, it can be taken as constant.