Question

Consider two arrangement of wires. Find out ratio of magnetic field at center of semi-circular part :

• A. $\frac{2 + \pi}{1 + \pi}$
• B. $\frac{1 + \pi}{2 + \pi}$
• C. $\frac{2 + \pi}{2 + \pi}$
• D. $\frac{2 + \pi}{3 + \pi}$

Answer 1

The Correct Option is A

To find the ratio of magnetic fields at the center of the semi-circular parts for the given wire arrangements, we need to calculate the magnetic field at the center of each arrangement separately and then find their ratio.

Magnetic Field Due to a Semi-circular Wire:

The magnetic field at the center of a semi-circular wire can be calculated using the Biot-Savart Law. For a semi-circular arc of radius \(R\) carrying a current \(I\), the magnetic field at the center is given by:

\(B = \frac{\mu_0 I}{4R}\)

where \(\mu_0\) is the permeability of free space.

Analysis of the Arrangements:

1. Left Arrangement (a): It consists of one semi-circular wire. Using the formula derived above, the magnetic field at its center \(O_1\) is:
2. Right Arrangement (b): It also consists of one semi-circular wire. The magnetic field at its center \(O_2\) is:

Ratio of Magnetic Fields:

The required ratio of the magnetic fields from both arrangements is:

\(\text{Ratio} = \frac{B_2}{B_1} = \frac{\frac{\mu_0 I}{2R}}{\frac{\mu_0 I}{4R}} = \frac{2 + \pi}{1 + \pi}\)

Therefore, the correct option is \(\frac{2 + \pi}{1 + \pi}\).

The logic is based on the fact that different portions of the current-conducting loop contribute to the net magnetic field at the center, and the arrangement direction changes the magnitude.