Question

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) $\dfrac{{{\mu _0}{I_1}{I_2}}}{{4\pi l}}\log \left( {dl} \right)$

Answer 1

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.