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Moving Charges and Magnetism — Physics STD 12 Science — Question

Bihar BoardEnglish MediumSTD 12 SciencePhysicsMoving Charges and Magnetism5 Marks
Question
A rectangular current carrying loop is placed 1cm 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 F1 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 × 10-4N towards left $\Big(\frac{\mu_0}{2\pi}=10^{-7}\Big)$

Force F2 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 ×10-4N towards right.

Hence, net force on the loop = 2.72N towards left.

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