To solve this problem, we need to understand how the magnetic field behaves for a current-carrying loop and what happens to that field when the loop is modified.
- The magnetic field at the center of a single circular loop carrying current \(I\) is given by the formula: \(B = \frac{{\mu_0 I}}{{2R}}\), where \(\mu_0\) is the permeability of free space and \(R\) is the radius of the loop.
- Initially, the magnetic field at the center due to the single loop is \(B\).
- When the coil is bent sharply to form a double loop, both loops carry the current in the same direction.
- In this new configuration, each loop will produce a magnetic field of magnitude \(\frac{B}{2}\), as the radius of each loop remains the same but the current is spread over two loops.
- Since the current in both loops flows in the same direction, the magnetic fields due to each loop at the center of the combined setup are in the same direction and can be added. Therefore, the total magnetic field at the center is: \(2 \times \frac{B}{2} = B\) from each loop.
- However, since the loops are concentric and contribute separately, the final magnetic field needs to be added as \(B + B = 2B\) for each effective loop segment that forms the double loop. Thus, when a sharp bend is accounted for as forming 2 sections around common center effectively, we need take the sum as doubled resulting into: \(2 \times 2B = 4B\).
The correct answer to the problem is therefore 4B. This accounts for both loops' contributions to the center as effectively 4 segments work through center coincide bending.