A circular coil having `N` turns is made from a wire of length `L` meter. If a current `I` ampere is passed through it and is placed in amagnetic field of `B` Tesla, the maximum torque on it is

Options

A. Directly proportional to N
B. Inversely proportional to N
C. Inversely proportional to `N^(2)`
D. Independent of N

Answer

To solve the problem, we need to determine the maximum torque (\( \tau_{\text{max}} \)) on a circular coil with \( N \) turns, made from a wire of length \( L \), carrying a current \( I \) in a magnetic field \( B \).

### Step-by-Step Solution:

1. **Understanding Torque in a Magnetic Field**:
The torque (\( \tau \)) experienced by a current-carrying coil in a magnetic field is given by the formula:
\[ \tau = m \cdot B \cdot \sin \theta \]
where \( m \) is the magnetic moment, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the magnetic moment and the magnetic field.

2. **Maximum Torque Condition**:
The maximum torque occurs when \( \sin \theta = 1 \), which means \( \theta = 90^\circ \). Therefore, the maximum torque can be expressed as:
\[ \tau_{\text{max}} = m \cdot B \]

3. **Calculating Magnetic Moment**:
The magnetic moment \( m \) for a coil with \( N \) turns, carrying a current \( I \), and having an area \( A \) is given by:
\[ m = N \cdot I \cdot A \]
For a circular coil, the area \( A \) can be expressed in terms of the radius \( r \):
\[ A = \pi r^2 \]
Thus, we can write:
\[ m = N \cdot I \cdot \pi r^2 \]

4. **Finding the Radius**:
The wire of length \( L \) forms \( N \) turns, and the circumference of one turn is \( 2\pi r \). Therefore, we have:
\[ N \cdot (2\pi r) = L \]
From this, we can solve for \( r \):
\[ r = \frac{L}{2\pi N} \]

5. **Substituting the Radius into the Magnetic Moment**:
Now substituting \( r \) back into the equation for \( m \):
\[ A = \pi \left(\frac{L}{2\pi N}\right)^2 = \pi \cdot \frac{L^2}{4\pi^2 N^2} = \frac{L^2}{4\pi N^2} \]
Therefore, the magnetic moment becomes:
\[ m = N \cdot I \cdot \frac{L^2}{4\pi N^2} = \frac{I L^2}{4\pi N} \]

6. **Calculating Maximum Torque**:
Now substituting \( m \) back into the equation for maximum torque:
\[ \tau_{\text{max}} = m \cdot B = \left(\frac{I L^2}{4\pi N}\right) \cdot B = \frac{I L^2 B}{4\pi N} \]

### Final Result:
Thus, the maximum torque on the coil is given by:
\[ \tau_{\text{max}} = \frac{I L^2 B}{4\pi N} \]