Current through ABC and A^{prime} B^{prime} C | vedclass

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(B) The magnetic field at point $P$ is produced by the segments $BC$ and $B^{\prime} C^{\prime}$.Segments $AB$ and $A^{\prime} B^{\prime}$ are directe

Vedclass · Accepted Answer

$B = \frac{\mu_0}{4 \pi} \left( \frac{2 I}{r} \right)$

Vedclass · Answer

(B) The magnetic field at point $P$ is produced by the segments $BC$ and $B^{\prime} C^{\prime}$.Segments $AB$ and $A^{\prime} B^{\prime}$ are directed towards and away from the axis containing $P$,so the magnetic field due to these segments at point $P$ is zero because $P$ lies on the line of these current elements.For the semi-infinite wire $BC$,the magnetic field at distance $r$ from the end $B$ is given by $B_1 = \frac{\mu_0 I}{4 \pi r}$.Similarly,for the semi-infinite wire $B^{\prime} C^{\prime}$,the magnetic field at distance $r$ from the end $B^{\prime}$ is $B_2 = \frac{\mu_0 I}{4 \pi r}$.Using the right-hand rule,both fields at $P$ point in the same direction (into the page).Therefore,the total magnetic field is $B = B_1 + B_2 = \frac{\mu_0 I}{4 \pi r} + \frac{\mu_0 I}{4 \pi r} = \frac{\mu_0}{4 \pi} \left( \frac{2 I}{r} \right)$.

Current through $ABC$ and $A^{\prime} B^{\prime} C^{\prime}$ is $I$. What is the magnetic field at $P$? $BP = PB^{\prime} = r$. (Here $C^{\prime} B^{\prime} PBC$ are collinear).

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