A square loop of side l iskept in a uniform magnetic field B such that its plane makes an angle `alpha` with `vecB`. The loop carries a current i. The torque experienced by the loop in this position is

Options

A. `Bi l^(2)`
B. `Bil^(2)sin alpha`
C. `Bil^(2) cos alpha`
D. zero

Answer

To find the torque experienced by a square loop of side length \( l \) carrying a current \( i \) in a uniform magnetic field \( B \) at an angle \( \alpha \) with respect to the magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Magnetic Moment**:
The magnetic moment \( \vec{m} \) of a current-carrying loop is given by the formula:
\[ \vec{m} = n \cdot i \cdot A \]
where \( n \) is the number of turns (which is 1 for a single loop), \( i \) is the current, and \( A \) is the area of the loop. For a square loop of side \( l \):
\[ A = l^2 \]
Therefore, the magnetic moment becomes:
\[ \vec{m} = i \cdot l^2 \]

2. **Determine the Angle Between Magnetic Moment and Magnetic Field**:
The angle \( \theta \) between the magnetic moment \( \vec{m} \) and the magnetic field \( \vec{B} \) is given by:
\[ \theta = 90^\circ - \alpha \]
This is because the plane of the loop makes an angle \( \alpha \) with the magnetic field.

3. **Calculate the Torque**:
The torque \( \vec{\tau} \) experienced by the loop in a magnetic field is given by the cross product:
\[ \vec{\tau} = \vec{m} \times \vec{B} \]
The magnitude of the torque can be expressed as:
\[ \tau = m \cdot B \cdot \sin(\theta) \]
Substituting the expression for \( \theta \):
\[ \tau = m \cdot B \cdot \sin(90^\circ - \alpha) = m \cdot B \cdot \cos(\alpha) \]

4. **Substituting the Magnetic Moment**:
Now, substituting the value of \( m \):
\[ \tau = (i \cdot l^2) \cdot B \cdot \cos(\alpha) \]

5. **Final Expression for Torque**:
Thus, the torque experienced by the square loop is:
\[ \tau = i \cdot l^2 \cdot B \cdot \cos(\alpha) \]

### Final Answer:
The torque experienced by the loop is:
\[ \tau = i \cdot l^2 \cdot B \cdot \cos(\alpha) \]