A circular coil carrying a current I has radius R and number of turns N. If all the three, i.e., the current I, radius R and number of turns N are doubled, then, magnetic, field at its centre becomes:

Options

A. Double
B. Half
C. Four times
D. One-fourth

Answer

To determine the effect on the magnetic field at the center of a circular coil when the current, radius, and number of turns are all doubled, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Magnetic Field**:
The magnetic field \( B \) at the center of a circular coil is given by the formula:
\[ B = \frac{\mu_0 N I}{2R} \]
where:
- \( B \) is the magnetic field,
- \( \mu_0 \) is the permeability of free space,
- \( N \) is the number of turns,
- \( I \) is the current,
- \( R \) is the radius of the coil.

2. **Identify the Changes**:
According to the problem, we need to double the current \( I \), the radius \( R \), and the number of turns \( N \):
- New current \( I' = 2I \)
- New radius \( R' = 2R \)
- New number of turns \( N' = 2N \)

3. **Substitute the New Values into the Formula**:
We substitute \( I' \), \( R' \), and \( N' \) into the magnetic field formula:
\[ B' = \frac{\mu_0 N' I'}{2R'} \]
Substituting the new values:
\[ B' = \frac{\mu_0 (2N)(2I)}{2(2R)} \]

4. **Simplify the Expression**:
Now simplify the expression:
\[ B' = \frac{\mu_0 (4NI)}{4R} \]
This simplifies to:
\[ B' = \frac{\mu_0 N I}{R} \]

5. **Compare with the Original Magnetic Field**:
The original magnetic field \( B \) was:
\[ B = \frac{\mu_0 N I}{2R} \]
Now we see that:
\[ B' = 2 \left(\frac{\mu_0 N I}{2R}\right) = 2B \]

6. **Conclusion**:
Therefore, the new magnetic field \( B' \) is twice the original magnetic field \( B \). Thus, the magnetic field at the center of the coil becomes doubled.

### Final Answer:
The magnetic field at the center becomes **doubled**.

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