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Moving Charges and Magnetism — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsMoving Charges and Magnetism5 Marks
Question
A rectangular current carrying loop is placed $1\ cm$ away from a long straight current$-$carrying conductor as shown.
  1. Will the net force acting on the loop due to straight conductor be attractive or repulsive in nature? Justify your answer. Calculate the magnitude of this force.
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Answer

The net force acting on the loop will be repulsive in nature.Justification:
Part $AB$ of the loop will experience a force of repulsion whereas part $CD$ will experience attraction. Parts $BC$ and $AD$ will not experience any force. Thus, the overall force will be a force of repulsion because $AB$ is closer to the straight conductor than $CD$ and the force between two current carrying conductors is inversely proportional to the distance between them.
$\text{F}=\frac{\mu_0}{2\pi}\frac{\text{I}_1\text{I}_2}{\text{r}}\text{l}$
Net Force
Force $F_1$ on $AB$ due to current in the straight conductor.
$\text{F}_1=\frac{\mu_0}{2\pi}\times\frac{2\times10\times15}{1\times10^{-2}}\times20\times10^{-2}$
$= 3 \times 10^{-4}N $towards left $\Big(\frac{\mu_0}{2\pi}=10^{-7}\Big)$
Force $F_2$ on $CD$ due to current in thw straight conductor.
$\text{F}_2=\frac{\mu_0}{2\pi}\times\frac{2\times10\times15}{11\times10^{-2}}\times20\times10^{-2}$
$= 0.2725 \times 10^{-4}N$ towards right.
Hence, net force on the loop $= 2.72N$ towards left.

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