Question

The magnetic field at the centre of a current loop of radius \(r\) carrying a current \(I\) is

• A. \(\frac{\mu_0 I}{2r}\)
• B. \(\frac{\mu_0 I}{r}\)
• C. \(\frac{\mu_0 I}{\pi r}\)
• D. \(\frac{2\mu_0 I}{r}\)
• E. \(\frac{\mu_0 I}{2\pi r}\)

Hint:

For a circular arc subtending an angle \(\theta\) at the center, \(B = \frac{\mu_0 I}{4\pi r} \theta\) (with \(\theta\) in radians). For a full loop, \(\theta = 2\pi\), giving \(B = \frac{\mu_0 I}{2r}\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This question asks for the standard formula for the magnetic field (\(B\)) produced at the very center of a circular loop of wire carrying a current (\(I\)). This formula is a direct result of applying the Biot-Savart Law.
Step 2: Key Formula or Approach:
The Biot-Savart Law gives the magnetic field \(d\vec{B}\) produced by a small current element \(I d\vec{l}\):
\[ d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times \hat{r}}{r^2} \] To find the total magnetic field at the center of a circular loop, we need to integrate this expression over the entire loop.
Step 3: Detailed Explanation:
For a circular loop of radius \(r\), the current element \(d\vec{l}\) is always perpendicular to the vector \(\vec{r}\) pointing from the element to the center of the loop. Therefore, the angle between \(d\vec{l}\) and \(\hat{r}\) is 90°, and \(\sin(90^\circ) = 1\).
The magnitude of the magnetic field from a small element \(dl\) is:
\[ dB = \frac{\mu_0}{4\pi} \frac{I dl}{r^2} \] By the right-hand rule, the direction of the magnetic field from every element \(dl\) on the loop points along the axis, perpendicular to the plane of the loop. So, we can simply sum up the magnitudes.
To find the total magnetic field \(B\), we integrate \(dB\) around the entire loop:
\[ B = \int_{\text{loop}} dB = \int_0^{2\pi r} \frac{\mu_0}{4\pi} \frac{I}{r^2} dl \] Since \(\frac{\mu_0 I}{4\pi r^2}\) is constant, we can take it out of the integral:
\[ B = \frac{\mu_0 I}{4\pi r^2} \int_0^{2\pi r} dl \] The integral of \(dl\) over the entire loop is just the circumference of the loop, which is \(2\pi r\).
\[ B = \frac{\mu_0 I}{4\pi r^2} (2\pi r) \] Simplifying the expression:
\[ B = \frac{\mu_0 I}{2r} \] Step 4: Final Answer:
This is a standard result. The magnetic field at the centre of a current loop of radius \(r\) carrying a current \(I\) is \(\frac{\mu_0 I}{2r}\), which corresponds to option (A).