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 \( l \) carrying a current \( i \) in a uniform magnetic field \( B \) at an angle \( \alpha \), we can follow these steps:

### Step 1: Understand the Magnetic Moment
The magnetic moment \( \vec{M} \) of a current-carrying loop is given by the formula:
\[ \vec{M} = n \cdot i \cdot \vec{A} \]
where:
- \( n \) is the number of turns (for a single loop, \( n = 1 \)),
- \( i \) is the current flowing through the loop,
- \( \vec{A} \) is the area vector of the loop, which is perpendicular to the plane of the loop.

For a square loop of side \( l \), the area \( A \) is:
\[ A = l^2 \]
Thus, the magnetic moment becomes:
\[ \vec{M} = i \cdot l^2 \cdot \hat{n} \]
where \( \hat{n} \) is the unit vector perpendicular to the plane of the loop.

### Step 2: Determine the Torque
The torque \( \vec{\tau} \) experienced by a magnetic moment in a magnetic field is given by:
\[ \vec{\tau} = \vec{M} \times \vec{B} \]
The magnitude of the torque can be expressed as:
\[ \tau = M \cdot B \cdot \sin(\theta) \]
where \( \theta \) is the angle between the magnetic moment \( \vec{M} \) and the magnetic field \( \vec{B} \).

### Step 3: Find the Angle
In this case, the loop makes an angle \( \alpha \) with the magnetic field \( \vec{B} \). The angle \( \theta \) between the magnetic moment \( \vec{M} \) and the magnetic field \( \vec{B} \) will be:
\[ \theta = 90^\circ - \alpha \]

### Step 4: Substitute Values
Now, substituting the values into the torque equation:
\[ \tau = M \cdot B \cdot \sin(90^\circ - \alpha) \]
Using the identity \( \sin(90^\circ - \alpha) = \cos(\alpha) \), we get:
\[ \tau = M \cdot B \cdot \cos(\alpha) \]

### Step 5: Substitute the Magnetic Moment
Substituting \( M = i \cdot l^2 \):
\[ \tau = (i \cdot l^2) \cdot B \cdot \cos(\alpha) \]

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