Moving Charges and Magnetism - Result Question 59
60. A square loop $A B C D$ carrying a current $i$, is placed near and coplanar with a long straight conductor XY carrying a current I, the net force on the loop will be :
[2016]
(a) $\frac{2 \mu_0 I i}{3 \pi}$
(b) $\frac{\mu_0 I i}{2 \pi}$
(c) $\frac{2 \mu_0 I i L}{3 \pi}$
(d) $\frac{\mu_0 I i L}{2 \pi}$
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Answer:
Correct Answer: 60. (a)
Solution:
- (a) The direction of current in conductor
$XY$ and $AB$ is same
$\therefore \quad F _{AB}=i \ell B \quad$ (attractive)
$F _{AB}=i(L) \cdot \frac{\mu_0 I}{2 \pi(\frac{L}{2})}(arrow)=\frac{\mu_0 iI}{\pi}(arrow)$
$F _{BC}$ opposite to $F _{AD}$
$F _{BC}(\uparrow)$ and $F _{AD}(\downarrow)$
$\Rightarrow$ cancels each other
$F _{CD}=i \ell B$ (repulsive)
$F _{CD}=i(L) \frac{\mu_0 I}{2 \pi(\frac{3 L}{2})}(arrow)=\frac{\mu_0 i I}{3 \pi}(arrow)$
Therefore the net force on the loop
$F _{\text{net }}=F _{AB}+F _{BC}+F _{CD}+F _{AD}$
$\Rightarrow F _{\text{net }}=\frac{\mu_o iI}{\pi}-\frac{\mu_o iI}{3 \pi}=\frac{2 \mu_o iI}{3 \pi}$
