Question

What happens to the magnetic field at the center of a circular coil if the number of turns is doubled while keeping the current constant?

• A. It becomes half \( \left(\frac{B}{2}\right) \)
• B. It remains the same
• C. It doubles \( (2B) \)
• D. It becomes four times \( (4B) \)

Hint:

For a circular current loop, \[ B = \frac{\mu_0 n I}{2R} \] Magnetic field increases directly with the number of turns and current, and decreases with increasing radius.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are investigating how the magnetic induction \( B \) at the center of a loop changes when the physical parameter (number of turns \( n \)) is modified.
Step 2: Key Formula or Approach:
The magnetic field at the center of a circular coil of \( n \) turns is:
\[ B = \frac{\mu_0 n I}{2R} \]
Where \( I \) is the current and \( R \) is the radius.
Step 3: Detailed Explanation:
From the formula, we can see that if the current \( I \) and radius \( R \) are kept constant, the magnetic field is directly proportional to the number of turns:
\[ B \propto n \]
If the initial field is \( B_1 \) for \( n_1 \) turns and the new field is \( B_2 \) for \( n_2 \) turns:
\[ \frac{B_2}{B_1} = \frac{n_2}{n_1} \]
Given that the number of turns is doubled, \( n_2 = 2n_1 \).
\[ B_2 = B_1 \times \left(\frac{2n_1}{n_1}\right) = 2B_1 \]
Step 4: Final Answer:
The magnetic field doubles (\( 2B \)).