A triangular loop of side `l` carries a current `I`. 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. (a)Zero
B. (b)iBL
C. (c)`(sqrt(3))/(2)iB^(2)l^(2)`
D. (d)`(sqrt(3))/(4)iBl^(2)`

Answer

To solve the problem of finding 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: Determine the Area of the Triangular Loop
The area \( A \) of an equilateral triangle with side length \( l \) can be calculated using the formula:
\[ A = \frac{\sqrt{3}}{4} l^2 \]

### Step 2: Calculate the Magnetic Moment
The magnetic moment \( m \) of the loop is given by the product of the current \( I \) and the area \( A \):
\[ m = I \cdot A \]
Substituting the area we found in Step 1:
\[ m = I \cdot \frac{\sqrt{3}}{4} l^2 = \frac{\sqrt{3}}{4} I l^2 \]

### Step 3: Understand the Orientation of the Loop
Since the plane of the loop is in the direction of the magnetic field \( B \), the angle \( \theta \) between the magnetic moment \( m \) and the magnetic field \( B \) is \( 90^\circ \).

### Step 4: Calculate the Torque
The torque \( \tau \) on the loop in a magnetic field is given by the formula:
\[ \tau = m \cdot B \cdot \sin(\theta) \]
Since \( \theta = 90^\circ \), we have \( \sin(90^\circ) = 1 \):
\[ \tau = m \cdot B \]
Substituting the expression for \( m \) from Step 2:
\[ \tau = \left(\frac{\sqrt{3}}{4} I l^2\right) \cdot B \]
Thus, the final expression for the torque is:
\[ \tau = \frac{\sqrt{3}}{4} I B l^2 \]

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

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