Question

Two wires of same material A and B, having lengths \(l\) and \(\frac{2l}{3}\) carry the currents \(I\) and \(\frac{3I}{4}\) respectively. Both the wires (A and B) are bent into a circles (of single turn). If \(B_{A}\) and \(B_{B}\) are the values of magnetic field at the centres of the respective circles, then the ratio \(\frac{B_{B}}{B_{A}}\) is

• A. \(\frac{3}{4}\)
• B. \(\frac{2}{3}\)
• C. \(\frac{9}{8}\)
• D. \(\frac{3}{8}\)

Hint:

Each wire becomes a circle of radius l divided by 2 pi; the field at the centre is mu0 I over 2r.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
A wire of length \(l\) bent into one circular turn has radius \(r=\dfrac{l}{2\pi}\).

Step 2: Field at the centre
\(B=\dfrac{\mu_0I}{2r}=\dfrac{\mu_0I\pi}{l}\), so \(B\propto\dfrac{I}{l}\).

Step 3: Ratio
\(\dfrac{B_B}{B_A}=\dfrac{3I/4}{2l/3}\times\dfrac{l}{I}=\dfrac{3}{4}\times\dfrac{3}{2}\)

Step 4: Result
\(\dfrac{B_B}{B_A}=\dfrac{9}{8}\), option (C).

Final Answer:
The ratio \(\dfrac{B_B}{B_A}\) is \(\dfrac{9}{8}\). \[ \boxed{\text{(C) }\dfrac{9}{8}} \]