A triangular loop of side l carries a current l. It is placed in a magnetic field B such that the plane of the loop is in the direction of B. The torque on the loop is

Options

A. Zero
B. IBl
C. `sqrt(3)/2II^(2)B^(2)`
D. `sqrt(3)/4IBl^(2)`

Answer

To find the torque on a triangular loop of side \( l \) carrying a current \( I \) in a magnetic field \( B \), we can follow these steps:

### Step 1: Understand the Torque Formula
The torque \( \tau \) on a current-carrying loop in a magnetic field is given by the formula:
\[ \tau = n \cdot I \cdot A \cdot B \cdot \sin(\alpha) \]
where:
- \( n \) is the number of turns (in this case, \( n = 1 \)),
- \( I \) is the current,
- \( A \) is the area of the loop,
- \( B \) is the magnetic field strength,
- \( \alpha \) is the angle between the normal to the loop and the magnetic field.

### Step 2: Identify the Angle
Since the plane of the loop is in the direction of the magnetic field, the normal to the loop makes an angle of \( 90^\circ \) with the magnetic field. Therefore, \( \sin(\alpha) = \sin(90^\circ) = 1 \).

### Step 3: Calculate the Area of the Triangular Loop
The area \( A \) of an equilateral triangle with side length \( l \) is given by:
\[ A = \frac{\sqrt{3}}{4} l^2 \]

### Step 4: Substitute Values into the Torque Formula
Now substituting the values into the torque formula:
\[ \tau = 1 \cdot I \cdot \left(\frac{\sqrt{3}}{4} l^2\right) \cdot B \cdot 1 \]
This simplifies to:
\[ \tau = \frac{\sqrt{3}}{4} I B l^2 \]

### Final Answer
Thus, the torque on the triangular loop is:
\[ \tau = \frac{\sqrt{3}}{4} I B l^2 \]

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