Question

plz answer 20 and 21 st question❤​

Answer 1

Answer:

(20)  Option C

Curved conductor=$B_{c}=\frac{u_{o}I}{2r}×\frac{3}{4}=\frac{u_{o}I \pi}{4\pi r}=\frac{3}{2}$

Lower straight conductor=$B_{s}=\frac{u_{o}I}{4\pi r}=\frac{u_{o}I}{4\pi r}$

Total field, $B=\frac{u_{o}I}{4\pi r}(\frac{3\pi}{2}+1)$

(21)  Option A

$B_{1}=B_{3}=0$

$B_{2}=\frac{u_{o}}{4\pi}×\frac{\pi i}{R_{1}}$

$B_{4}=\frac{u_{o}}{4\pi}×\frac{\pi i}{R_{2}}$   As |B₂|>|B₄|

$So, B_{net}=B_{2}-B_{4}$

⇒$B_{net}=\frac{u_{o}i}{4}[\frac{1}{R_{1}}-\frac{1}{R_{2}}]$

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