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:

  1. (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

$F _{\text{net }}=F _{AB}+F _{BC}+F _{CD}+F _{AD}$

$\Rightarrow F _{\text{net }}=\frac{\mu_o iI}{\pi}-\frac{\mu_o iI}{3 \pi}=\frac{2 \mu_o iI}{3 \pi}$