A rectangular coil `20cmxx20cm` has `100` turns and carries a current of `1 A`. It is placed in a uniform magnetic field `B=0.5 T` with the direction of magnetic field parallel to the plane of the coil. The magnitude of the torque required to hold this coil in this position is

Options

A. Zero
B. 200N-m
C. 2N-m
D. 10N-m

Answer

To find the magnitude of the torque required to hold the rectangular coil in the given position, we can follow these steps:

### Step 1: Identify the parameters given in the problem
- Dimensions of the coil: 20 cm x 20 cm
- Number of turns (N): 100
- Current (I): 1 A
- Magnetic field (B): 0.5 T

### Step 2: Convert the dimensions of the coil into meters
Since the dimensions are given in centimeters, we need to convert them into meters:
- Length of one side of the coil = 20 cm = 0.2 m
- Area (A) of the coil = Length x Width = 0.2 m x 0.2 m = 0.04 m²

### Step 3: Calculate the magnetic moment (M) of the coil
The magnetic moment (M) is given by the formula:
\[ M = N \times I \times A \]
Substituting the values:
\[ M = 100 \times 1 \, \text{A} \times 0.04 \, \text{m}^2 \]
\[ M = 4 \, \text{A m}^2 \]

### Step 4: Determine the angle between the magnetic moment and the magnetic field
The problem states that the magnetic field is parallel to the plane of the coil. Therefore, the angle (θ) between the magnetic moment vector and the magnetic field vector is 90 degrees (90°).

### Step 5: Calculate the torque (τ)
The torque (τ) can be calculated using the formula:
\[ \tau = M \times B \times \sin(\theta) \]
Since θ = 90°, sin(90°) = 1, we can simplify the equation to:
\[ \tau = M \times B \]
Substituting the values:
\[ \tau = 4 \, \text{A m}^2 \times 0.5 \, \text{T} \]
\[ \tau = 2 \, \text{N m} \]

### Step 6: Conclusion
The magnitude of the torque required to hold the coil in this position is **2 N·m**.

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