Question
A rectangular loop of sides 25cm and 10cm carrying a current of 15A is placed with its longer side parallel to a long straight conductor 2.0cm apart carrying a current of 25A (fig). What is the net force on the loop?
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Answer
Rectangular loop PQRS is placed near a long straight wire as shown.
$l = PQ = RS = 25cm = 0.25m, b = QR = PS = 10cm = 0.10m, r_1 = 2cm = 0.02m, r_2 = 12cm = 0.12m,$
The currents in PQ and XY are antiparallel, so PQ is repelled away from wire XY. This repulsive force is: $\text{F}_1=\frac{\mu_0\text{I}_1\text{I}_2}{2\pi\text{r}_1}\text{l}$ $=\frac{4\pi\times10^{-7}\times25\times15\times0.25}{2\pi\times0.02}=9.375\times10^{-4}\text{N}$
The currents in XY and RS are in the same direction,
so wire RS is attracted towards wire XY.
This attractive force is:$\text{F}_1=\frac{\mu_0\text{I}_1\text{I}_2}{2\pi\text{r}_1}\text{l}$
$=\frac{4\pi\times10^{-7}\times25\times15\times0.25}{2\pi\times0.12}$
$=1.563\times10^{-4}\text{N}$ The currents in PS and QR are equal and opposite.
By symmetry they exert equal and opposite forces (F3 and F4) and hence net force on these sides is zero.
$\therefore$ Net force on rectangular loop, $F = F_1 – F_2$ (repulsive)
$= 9.375 \times 10^{–4} – 1.563 \times 10^{–4} = 7.812 \times 10^{–4} N$ (repulsive)
The net force is directed away from long wire XY.
$l = PQ = RS = 25cm = 0.25m, b = QR = PS = 10cm = 0.10m, r_1 = 2cm = 0.02m, r_2 = 12cm = 0.12m,$
The currents in PQ and XY are antiparallel, so PQ is repelled away from wire XY. This repulsive force is: $\text{F}_1=\frac{\mu_0\text{I}_1\text{I}_2}{2\pi\text{r}_1}\text{l}$ $=\frac{4\pi\times10^{-7}\times25\times15\times0.25}{2\pi\times0.02}=9.375\times10^{-4}\text{N}$
The currents in XY and RS are in the same direction,
so wire RS is attracted towards wire XY.
This attractive force is:$\text{F}_1=\frac{\mu_0\text{I}_1\text{I}_2}{2\pi\text{r}_1}\text{l}$
$=\frac{4\pi\times10^{-7}\times25\times15\times0.25}{2\pi\times0.12}$
$=1.563\times10^{-4}\text{N}$ The currents in PS and QR are equal and opposite.
By symmetry they exert equal and opposite forces (F3 and F4) and hence net force on these sides is zero.
$\therefore$ Net force on rectangular loop, $F = F_1 – F_2$ (repulsive)
$= 9.375 \times 10^{–4} – 1.563 \times 10^{–4} = 7.812 \times 10^{–4} N$ (repulsive)
The net force is directed away from long wire XY.
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