Question

A circular loop A of radius \( R \) carries a current \( I \). Another circular loop B of radius \( r = \frac{R}{20} \) is placed concentrically in the plane of A. The magnetic flux linked with loop B is proportional to:

• A. R
• B. \(√R\)
• C. \(R^{\frac{3}{2}}\)
• D. \(R^{2}\)

Hint:

In the case of two concentric loops, the magnetic flux linked with the smaller loop is proportional to the radius of the larger loop.

Answer 1

The Correct Option is B

Magnetic Flux Linked with Loop B 
Step 1: Magnetic Field Due to Loop A
- The magnetic field \( B \) at the center of a circular loop carrying a current \( I \) is given by the formula: \[ B = \frac{\mu_0 I}{2R} \] where:
- \( R \) = radius of loop A,
- \( \mu_0 \) = permeability of free space. 

Step 2: Magnetic Flux Linked with Loop B
- The magnetic flux \( \Phi_B \) linked with loop B is given by: \[ \Phi_B = B \times A_B \] where:
- \( A_B = \pi r^2 \) is the area of loop B,
- \( r = \frac{R}{20} \) is the radius of loop B. Thus, \[ \Phi_B = \left( \frac{\mu_0 I}{2R} \right) \times \pi \left( \frac{R}{20} \right)^2 \] Simplifying: \[ \Phi_B = \frac{\mu_0 I \pi R^2}{2R \times 400} = \frac{\mu_0 I \pi R}{800} \] Thus, the magnetic flux linked with loop B is directly proportional to \( R \). 

Step 3: Conclusion
The magnetic flux linked with loop B is proportional to \( R \), so the correct answer is: \[ \boxed{(A) \, R} \]