A loop of area 'S' `m^2` and N turns carrying current 'i' is placed in a uniform magnetic field 'B' with its plane parallel to `vecB`. If torque `tau` is experienced by loop due to magnetic field with its plane parallel to `vecB`. If torque `tau` is experienced by loop due to magnetic field find I`vecB` I

Options

A. `tau/ (NiS)`
B. `(Ntau)/(iS)`
C. `(itau)/(NS)`
D. `(Stau)/(Ni)`

Answer

To solve the problem, we need to find the magnetic field \( B \) experienced by a loop of area \( S \) with \( N \) turns carrying current \( I \), placed in a uniform magnetic field with its plane parallel to \( \vec{B} \). The torque \( \tau \) experienced by the loop is given by the equation for torque on a current loop in a magnetic field.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Area of the loop: \( S \) (in \( m^2 \))
- Number of turns: \( N \)
- Current flowing through the loop: \( I \)
- Torque experienced by the loop: \( \tau \)

2. **Understand the Magnetic Moment**:
The magnetic moment \( m \) of a loop is given by the formula:
\[ m = N \cdot I \cdot A \]
where \( A \) is the area of the loop. In this case, \( A = S \), so:
\[ m = N \cdot I \cdot S \]

3. **Torque on the Loop**:
The torque \( \tau \) experienced by a current loop in a magnetic field is given by:
\[ \tau = m \cdot B \cdot \sin(\theta) \]
where \( \theta \) is the angle between the magnetic moment \( m \) and the magnetic field \( B \). Since the plane of the loop is parallel to \( \vec{B} \), the angle \( \theta = 90^\circ \). Thus, \( \sin(90^\circ) = 1 \), and the equation simplifies to:
\[ \tau = m \cdot B \]

4. **Substituting the Magnetic Moment**:
Substitute the expression for the magnetic moment \( m \) into the torque equation:
\[ \tau = (N \cdot I \cdot S) \cdot B \]

5. **Rearranging to Find \( B \)**:
To find the magnetic field \( B \), rearrange the equation:
\[ B = \frac{\tau}{N \cdot I \cdot S} \]

### Final Answer:
Thus, the magnetic field \( B \) is given by:
\[ B = \frac{\tau}{N \cdot I \cdot S} \]