Question

An infinitely long straight conductor is bent into the shape as shown below. It carries a current of \(1\) ampere and the radius of the circular loop is \(R\) metre. Then, the magnitude of magnetic induction at the centre of the circular loop is

• A. \(\dfrac{\mu_0 I}{2\pi R}\)
• B. \(\dfrac{\mu_0 \pi I}{2R}\)
• C. \(\dfrac{\mu_0 I}{2\pi R}(\pi + 1)\)
• D. \(\dfrac{\mu_0 I}{2\pi R}(\pi - 1)\)

Hint:

Magnetic field contributions add vectorially. A full circular loop gives \(\frac{\mu_0 I}{2R}\) and infinite straight wire gives \(\frac{\mu_0 I}{2\pi R}\).

Answer 1

The Correct Option is C

Step 1: Field at centre due to circular loop.
For a full circular loop:
\[ B_{loop} = \frac{\mu_0 I}{2R} \]
Step 2: Field due to infinitely long straight wire part.
Magnetic field at distance \(R\) from an infinite straight wire:
\[ B_{wire} = \frac{\mu_0 I}{2\pi R} \]
Step 3: Total field at centre.
Both contributions are in same direction, so:
\[ B_{total} = B_{loop} + B_{wire} \]
Step 4: Express in common form.
\[ B_{loop} = \frac{\mu_0 I}{2R} = \frac{\mu_0 I}{2\pi R}\cdot \pi \]
So:
\[ B_{total} = \frac{\mu_0 I}{2\pi R}(\pi + 1) \]
Final Answer:
\[ \boxed{\dfrac{\mu_0 I}{2\pi R}(\pi + 1)} \]