MCQ
The resultant force on the current loop $PQRS$ due to a long current carrying conductor will be
- A$10^{-4}\,N$
- B$3.6 \times {10^{ - 4}}\,N$
- C$1.8 \times {10^{ - 4}}\,N$
- ✓$5 \times {10^{ - 4}}\,N$
✓
Answer
Correct option: D.
$5 \times {10^{ - 4}}\,N$
d
Force on $SR$ and $PQ$ are equal but opposite so their net force will be zero. Force between two parallel conductors carrying currents $\mathrm{I}_{1}$ and $\mathrm{I}_{2}$
Force on $SR$ and $PQ$ are equal but opposite so their net force will be zero. Force between two parallel conductors carrying currents $\mathrm{I}_{1}$ and $\mathrm{I}_{2}$
$\mathrm{F}=\frac{\mu_{0}}{2 \pi} \frac{\mathrm{I}_{1} \mathrm{I}_{2} \ell}{\mathrm{r}}$
Where,
$\mathrm{r}=$ distance between two parallel conductors
$F_{R} =\frac{\mu_{0} I_{1} I_{2}}{2 \pi}\left(\frac{1}{2}-\frac{1}{12}\right) \times \frac{1}{10^{-2}} \times 15 \times 10^{-2}$
$=2 \times 10^{-7} \times 20 \times 20 \times \frac{5}{12} \times 15$
$=5 \times 10^{-4} \,\mathrm{N}$
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