Question

A current carrying wire is first bent in the form of a circular loop and then bent in the form of a square loop. The ratio of the magnetic fields induced at the centres of the loops in the two cases is:

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

Hint:

- Use Ampere’s Law and Biot-Savart’s Law to derive expressions for the magnetic field at the centre of loops. - The wire length remains the same, so equate perimeter equations to find new dimensions. - The ratio of the fields follows from direct substitution and simplification.

Answer 1

The Correct Option is B

Step 1: Magnetic Field at the Centre of a Circular Loop For a circular loop of radius \( R \) carrying current \( I \), the magnetic field at its centre is given by: \[ B_{\text{circle}} = \frac{\mu_0 I}{2R} \] Since the length of the wire remains constant, \[ 2\pi R = L \] which gives \[ R = \frac{L}{2\pi} \] Substituting this in \( B_{\text{circle}} \): \[ B_{\text{circle}} = \frac{\mu_0 I}{2 \times \frac{L}{2\pi}} = \frac{\mu_0 I \pi}{L} \] Step 2: Magnetic Field at the Centre of a Square Loop For a square loop, each side is of length \( \frac{L}{4} \), and the magnetic field at the centre of a square loop is given by: \[ B_{\text{square}} = \frac{2\sqrt{2} \mu_0 I}{\pi a} \] where \( a \) is the side length of the square: \[ a = \frac{L}{4} \] Substituting this in \( B_{\text{square}} \): \[ B_{\text{square}} = \frac{2\sqrt{2} \mu_0 I}{\pi \times \frac{L}{4}} = \frac{8\sqrt{2} \mu_0 I}{\pi L} \] Step 3: Finding the Ratio The ratio of the magnetic fields is: \[ \frac{B_{\text{circle}}}{B_{\text{square}}} = \frac{\frac{\mu_0 I \pi}{L}}{\frac{8\sqrt{2} \mu_0 I}{\pi L}} \] Simplifying, \[ \frac{B_{\text{circle}}}{B_{\text{square}}} = \frac{\pi^2}{8\sqrt{2}} \] Thus, the correct answer is \( \frac{\pi^2}{8\sqrt{2}} \).