A circular loop has radius $R$ and carries current $I$ as shown in figure. In order that the net magnetic field at the centre of the loop is zero, the current in wire AB should have magnitude:
Show Hint
Equate \(\mu_0 I/(2R)\) with \(\mu_0 I'/(4\pi R)\) to get \(I' = 2\pi I\). The loop field points into the page, so the wire field must point out of the page, which needs current along +X.
Step 1: Plan:
I use the superposition of the two fields. I find the direction of each field first, then equate the sizes.
Step 2: Directions by the Right-Hand Rule:
The loop current in the figure is clockwise. Curl the fingers of the right hand along the current. The thumb points into the page, so $\vec{B}_{loop}$ at the centre is along $-\hat{z}$.
For the cancellation, $\vec{B}_{wire}$ at the centre must be along $+\hat{z}$ (out of the page).
Point the thumb of the right hand along the current in the wire. At a point above the wire, the fingers curl out of the page when the thumb points to the right (+X). So the current in AB must flow along $+X$.
Step 3: Sizes of the Two Fields:
$B_{loop} = \dfrac{\mu_0 I}{2R}$ and $B_{wire} = \dfrac{\mu_0 I'}{2\pi d}$ with $d = 2R$.
Step 4: Solve for the Current:
Set $B_{wire} = B_{loop}$: $\dfrac{\mu_0 I'}{4\pi R} = \dfrac{\mu_0 I}{2R}$.
Cancel $\mu_0$ and $R$: $\dfrac{I'}{4\pi} = \dfrac{I}{2}$, so $I' = 2\pi I$.
Step 5: Check the Options:
The size $2\pi I$ appears in options (A) and (B). Of these, only (A) has the +X direction that we found. The size $\pi I$ in (C) and (D) gives half the field needed.
Final Answer:
$I' = 2\pi I$ along the $+X$ axis, option (A).
\[ \boxed{2\pi I,\ \text{along } +X} \]
