An infinitely long straight conductor carrying current 'I' is bent into a shape as shown in figure. The radius of the circular loop is 'r'. The magnetic induction at the centre of the loop at point 'O' is ______.
Show Hint
To factor an expression like $\frac{X}{2r} \pm \frac{X}{2\pi r}$, pulling out $\frac{X}{2\pi r}$ will always leave you with $(\pi \pm 1)$ inside the brackets. This is a very common algebraic trick in electromagnetism problems!
Step 1: Understanding the Question:
An infinitely long wire carries a current but features a circular loop in the middle. We must find the net magnetic field at the very center of this loop by vectorially adding the fields produced by the straight wire sections and the circular loop section. Step 2: Detailed Explanation:
The total magnetic induction ($B_{\text{net}}$) at the center 'O' is the superposition of two components:
1. $B_{\text{straight}}$: The magnetic field created by the infinitely long straight wire.
2. $B_{\text{loop}}$: The magnetic field created by the full circular loop. Magnitude of components:
The magnetic field at distance $r$ from an infinite straight wire is:
$B_{\text{straight}} = \frac{\mu_0 I}{2\pi r}$
The magnetic field at the center of a full circular loop of radius $r$ is:
$B_{\text{loop}} = \frac{\mu_0 I}{2r}$ Direction of components:
To determine whether these fields add together or subtract, we apply the Right-Hand Grip Rule.
- Assume the current in the long straight wire flows from left to right. Pointing the right thumb to the right, the fingers curl out of the page above the wire and into the page below the wire.
- Assuming the loop lies above the straight wire, the current must travel around the loop in a counter-clockwise direction to return to the straight path. Curling fingers counter-clockwise means the thumb points out of the page.
(If the loop was formed by twisting the wire backward on itself crossing over, current would be clockwise, thumb pointing IN. Without the figure, we rely on the options provided. The presence of ($\pi - 1$) in the options heavily dictates they oppose each other).
Let's assume standard opposite directions:
$B_{\text{net}} = B_{\text{loop}} - B_{\text{straight}}$
$B_{\text{net}} = \frac{\mu_0 I}{2r} - \frac{\mu_0 I}{2\pi r}$
Factor out the common terms $\frac{\mu_0 I}{2\pi r}$ to match the options:
$B_{\text{net}} = \frac{\mu_0 I}{2\pi r} \times \pi - \frac{\mu_0 I}{2\pi r} \times 1$
$B_{\text{net}} = \frac{\mu_0 I}{2\pi r} (\pi - 1)$ Step 3: Final Answer:
The magnetic induction is $\frac{\mu_0 I}{2\pi r} (\pi - 1)$, matching option (d).
