A circular loop of area `1cm^2` , carrying a current of 10 A , is placed in a magnetic field of 0.1 T perpendicular to the plane of the loop. The torque on the loop due to the magnetic field is

Options

A. Zero
B. `10^(-4) N m`
C. `10^(-2) N m`
D. 1 N m

Answer

To solve the problem of finding the torque on a circular loop in a magnetic field, we can follow these steps:

### Step 1: Identify the given values
- Area of the loop (A) = 1 cm² = \(1 \times 10^{-4} \, \text{m}^2\) (since \(1 \, \text{cm}^2 = 10^{-4} \, \text{m}^2\))
- Current (I) = 10 A
- Magnetic field (B) = 0.1 T
- Number of loops (n) = 1 (since it is a single loop)

### Step 2: Write the formula for torque
The torque (\( \tau \)) acting on a current-carrying loop in a magnetic field is given by the formula:
\[ \tau = n \cdot B \cdot I \cdot A \cdot \sin(\theta) \]
where:
- \( \tau \) = torque
- \( n \) = number of loops
- \( B \) = magnetic field strength
- \( I \) = current
- \( A \) = area of the loop
- \( \theta \) = angle between the magnetic field and the normal to the loop's surface

### Step 3: Determine the angle \( \theta \)
In this case, the magnetic field is perpendicular to the plane of the loop. Therefore, the angle \( \theta = 0^\circ \).

### Step 4: Calculate \( \sin(\theta) \)
Since \( \theta = 0^\circ \):
\[ \sin(0^\circ) = 0 \]

### Step 5: Substitute the values into the torque formula
Now, substituting the values into the torque formula:
\[ \tau = n \cdot B \cdot I \cdot A \cdot \sin(0^\circ) = 1 \cdot 0.1 \, \text{T} \cdot 10 \, \text{A} \cdot (1 \times 10^{-4} \, \text{m}^2) \cdot 0 \]

### Step 6: Simplify the expression
Since \( \sin(0^\circ) = 0 \):
\[ \tau = 0 \]

### Final Answer
The torque on the loop due to the magnetic field is:
\[ \tau = 0 \, \text{N m} \]

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