Question

Find magnetic field at point ‘O’ in the given figure.

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

Hint:

For a circular current loop, the magnetic field at the center is given by \( B = \frac{\mu_0 I}{2R} \). For a long straight wire, the magnetic field at a distance \( R \) is given by \( B = \frac{\mu_0 I}{2 \pi R} \).

Answer 1

The Correct Option is C

Step 1: Understand the setup.
The problem involves a circular current-carrying loop with a point \( O \) located at the center of the loop. The magnetic field at the center of a circular loop of current is calculated using Ampère’s circuital law or Biot-Savart law.
Step 2: Magnetic field due to a current-carrying loop.
The magnetic field at the center of a circular loop of current \( I \) and radius \( R \) is given by the formula: \[ B = \frac{\mu_0 I}{2R} \] This is the magnetic field at the center due to the current circulating in the loop.
Step 3: Additional considerations for the setup.
In the problem, the magnetic field at point \( O \) is influenced by both the current in the loop and the contribution from the straight current-carrying conductor extending to infinity. The contribution from the straight wire is calculated using Ampère’s law for a long straight current-carrying wire: \[ B_{\text{wire}} = \frac{\mu_0 I}{2 \pi R} \] The total magnetic field at point \( O \) combines the effects of both the loop and the wire.
Step 4: Combine the contributions.
The total magnetic field at point \( O \) is: \[ B = \frac{\mu_0 I}{2R} \left( 1 - \frac{1}{\pi} \right) \] where the first term corresponds to the magnetic field from the loop, and the second term accounts for the magnetic field contribution from the straight wire.