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: Understanding the Concept:
The loop (radius \(R\), current \(I\)) and the long straight wire AB (current \(I'\)) both produce a magnetic field at the centre of the loop. For zero net field, they must be equal in size and opposite in direction.
From the figure, the wire AB lies in the same plane as the loop, at a distance \(2R\) below the centre. The positive z-axis points out of the page towards the viewer.
Step 2: Field of the Loop at its Centre:
The loop current runs clockwise as seen in the figure. By the right-hand rule, the field at the centre points into the page (negative z).
\[ B_{loop} = \frac{\mu_0 I}{2R} \]
Step 3: Field of the Wire at the Centre:
The centre is at distance \(2R\) from the wire.
\[ B_{wire} = \frac{\mu_0 I'}{2\pi (2R)} = \frac{\mu_0 I'}{4\pi R} \]
Step 5: Direction of the Wire Current:
The wire's field at the centre must point out of the page (positive z), to cancel the loop's field.
Take the current along +X. For a point above the wire (in +Y direction), the field direction is \(\hat{x} \times \hat{y} = +\hat{z}\), out of the page. This is what we need.
If the current were along -X, the field at the centre would point into the page and add to the loop field. So the current must flow along +X.
Step 6: Why the Other Options Are Wrong:
Options (C) and (D) have magnitude \(\pi I\), which gives only half the required wire field. Option (B) has the right magnitude but the wrong direction.
Final Answer:
The current in AB is \(2\pi I\), along the +X axis. This is option (A).
\[ \boxed{2\pi I,\ \text{along } +X} \]
