Question

Consider a thin long insulator coated conducting wire carrying current \( I \). It is now wound once around an insulating thin disc of radius \( R \) to bring the wire back on the same side, as shown in the figure. The magnetic field at the center of the disc is equal to: 

• A. \( \frac{\mu_0 I}{2R} \)
• B. \( \frac{\mu_0 I}{4R} \left[ 3 + \frac{2}{\pi} \right] \)
• C. \( \frac{\mu_0 I}{4R} \left[ 1 + \frac{2}{\pi} \right] \)
• D. \( \frac{\mu_0 I}{2R} \left[ 1 + \frac{1}{\pi} \right] \)

Hint:

For a current-carrying coil, the magnetic field at the center is affected by the number of turns and the geometry of the loop or disc.

Answer 1

The Correct Option is B

Step 1: Understanding the magnetic field.
The magnetic field at the center of a circular loop carrying current \( I \) is given by the formula \( B = \frac{\mu_0 I}{2R} \). For a coil of multiple turns or a disc with a conducting wire wound around it, the field is modified by a factor that accounts for the geometry of the setup. The magnetic field at the center of the disc is given by \( \frac{\mu_0 I}{4R} \left[ 3 + \frac{2}{\pi} \right] \).
Step 2: Conclusion.
Thus, the correct answer is option (B).