A current I is flowing in a conductor PQRST a | vedclass

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Question

(A) The magnetic field at the centre $O$ is the sum of the magnetic fields due to the three parts: straight wire $PQ$, curved part $QRS$, and straight

Vedclass · Accepted Answer

$\frac{\mu_0 I}{4 \pi r}\left(\frac{3 \pi}{2}+1\right)(-\widehat{k})$

Vedclass · Answer

(A) The magnetic field at the centre $O$ is the sum of the magnetic fields due to the three parts: straight wire $PQ$, curved part $QRS$, and straight wire $ST$.
$1$. For the straight wire $PQ$: The point $O$ lies on the axis of the wire, so the magnetic field $B_{PQ} = 0$.
$2$. For the curved part $QRS$: The angle subtended at the centre is $	heta = 270^\circ = \frac{3\pi}{2} 	ext{ radians}$. The magnetic field is $B_{QRS} = \frac{\mu_0 I}{4\pi r} 	heta = \frac{\mu_0 I}{4\pi r} \left(\frac{3\pi}{2}ight)$. Using the right-hand rule, the direction is into the page $(-\widehat{k})$.
$3$. For the straight wire $ST$: The point $O$ is at a perpendicular distance $r$ from the wire. The wire extends from $S$ to infinity. The magnetic field is $B_{ST} = \frac{\mu_0 I}{4\pi r} (\sin 90^\circ + \sin 0^\circ) = \frac{\mu_0 I}{4\pi r}$. Using the right-hand rule, the direction is into the page $(-\widehat{k})$.
Total magnetic field $B = B_{PQ} + B_{QRS} + B_{ST} = 0 + \frac{\mu_0 I}{4\pi r} \left(\frac{3\pi}{2}ight) + \frac{\mu_0 I}{4\pi r} = \frac{\mu_0 I}{4\pi r} \left(\frac{3\pi}{2} + 1ight) (-\widehat{k})$.

$A$ current $I$ is flowing in a conductor $PQRST$ as shown in the figure. The radius of the curved path $QRS$ is $r$ and the length of the straight portions $PQ$ and $ST$ is very large. The magnetic field at the centre $O$ of the curved part is $(\mu_0 = \text{permeability of free space})$

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