Consider two current carrying conductors in x-y plane, carrying constant current $I_{1} a n d I_{2}$ respectively as shown in the figure.Now consider two arbitrary elementary current elements AB & CD of lengths $d l_{1} a n d d l_{2}$ on the loop-1 and loop-2 respectively. Let $d {\vec{F}}_{12}$ be the force which the current carrying element $I_{2} d {\vec{l}}_{2}$ exerts on the current carrying element $I_{1} d {\vec{l}}_{1} & {\vec{F}}_{12}$ represent the net force on 1^st loop by the 2^nd loop, developed due to interaction between current carrying loops. Similarly $d {\vec{F}}_{21}$ be the force which the current
carrying element $I_{1} d {\vec{l}}_{1}$ exerts on the current carrying element $I_{2} d {\vec{l}}_{2} & {\vec{F}}_{21}$ represent the net force on 2^nd loop by the 1^st loop, developed due to interaction between current carrying loops. Choose the correct option(s)

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$d {\vec{F}}_{12} + d {\vec{F}}_{21} = \vec{0}$

$d {\vec{F}}_{12} + d {\vec{F}}_{21} \neq \vec{0}$

${\vec{F}}_{12} + {\vec{F}}_{21} = \vec{0}$

${\vec{F}}_{12} + {\vec{F}}_{21} \neq \vec{0}$

answer is B, C.

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Detailed Solution

$d {\vec{B}}_{12} =$ Magnetic field at any point on the elementary element AB due to current carrying element $I_{2} d {\vec{l}}_{2}$ .
$d {\vec{B}}_{12} = \frac{μ_{0}}{4 π} (\frac{I_{2} d \vec{l} \times {\vec{r}}_{12}}{r_{12}^{3}})$
$d {\vec{F}}_{12} = I_{2} d {\vec{l}}_{1} \times d {\vec{B}}_{12} = (\frac{μ_{0}}{4 π}) \frac{I_{1} d {\vec{l}}_{1} \times (I_{2} d {\vec{l}}_{2} \times {\vec{r}}_{12}}{r_{12}^{3}}$
The force $d {\vec{F}}_{21}$ will be in the plane of loop and perpendicular to the current carrying element $I_{1} d {\vec{l}}_{1}$ (say along ${\hat{p}}_{12}$ )
Similarly $d {\vec{F}}_{21} = \frac{μ_{0}}{4 π} \frac{I_{2} d {\vec{l}}_{2} \times (I_{1} d {\vec{l}}_{1} \times {\vec{r}}_{21})}{r_{21}^{3}}$ [say along ${\hat{p}}_{21}$ ]
$∴ {\hat{p}}_{12}$ is not parallel to , in general
So, $d {\vec{F}}_{12} + d {\vec{F}}_{21} \neq \vec{0}$
But ${\vec{F}}_{12} = \frac{μ_{0} I_{1} I_{2}}{4 π} \int_{L_{1}}^{} \int_{L_{2}}^{} \frac{d {\vec{l}}_{1} \times (d {\vec{l}}_{2} \times {\vec{r}}_{12})}{r_{12}^{3}}$
${\vec{F}}_{21} = \int_{L_{2}}^{} \int_{L_{1}}^{} \frac{d {\vec{l}}_{2} \times (d {\vec{l}}_{1} \times {\vec{r}}_{21})}{r_{21}^{3}}$
Hence, ${\vec{F}}_{12} + {\vec{F}}_{21} = \vec{0}$