A square loop of sides `10 cm` carries a current of `10 A`.A uniform magnetic field of magnitude `0.20 T` exists parallel to one of the side of the loop.(a)What is the force acting on the loop? (b)What is the torque acting on the loop?

Explanation

To solve the problem step by step, we will calculate both the force acting on the loop and the torque acting on the loop.

### Given Data:
- Side of the square loop, \( A = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Current, \( I = 10 \, \text{A} \)
- Magnetic field, \( B = 0.20 \, \text{T} \)

### Part (a): Calculating the Force Acting on the Loop

1. **Identify the Length of the Side of the Loop**:
The length of one side of the square loop is given as \( 10 \, \text{cm} \), which we convert to meters:
\[ L = 0.1 \, \text{m} \]

2. **Use the Formula for Force**:
The force \( F \) acting on a current-carrying conductor in a magnetic field is given by:
\[ F = I \cdot B \cdot L \]
where:
- \( I \) is the current,
- \( B \) is the magnetic field strength,
- \( L \) is the length of the conductor in the magnetic field.

3. **Substituting the Values**:
Now, substituting the values into the formula:
\[ F = 10 \, \text{A} \cdot 0.20 \, \text{T} \cdot 0.1 \, \text{m} \]

4. **Calculating the Force**:
\[ F = 10 \cdot 0.20 \cdot 0.1 = 0.2 \, \text{N} \]

### Part (b): Calculating the Torque Acting on the Loop

1. **Use the Formula for Torque**:
The torque \( \tau \) acting on a current loop in a magnetic field is given by:
\[ \tau = N \cdot I \cdot A \cdot B \cdot \cos(\theta) \]
where:
- \( N \) is the number of turns (for a single loop, \( N = 1 \)),
- \( A \) is the area of the loop,
- \( \theta \) is the angle between the magnetic field and the normal to the plane of the loop.

2. **Calculate the Area of the Loop**:
The area \( A \) of the square loop is:
\[ A = L^2 = (0.1 \, \text{m})^2 = 0.01 \, \text{m}^2 \]

3. **Determine the Angle**:
Since the magnetic field is parallel to one side of the loop, the angle \( \theta \) between the magnetic field and the normal to the loop is \( 90^\circ \). Thus, \( \cos(90^\circ) = 0 \).

4. **Substituting the Values**:
Now substituting the values into the torque formula:
\[ \tau = 1 \cdot 10 \, \text{A} \cdot 0.01 \, \text{m}^2 \cdot 0.20 \, \text{T} \cdot \cos(0^\circ) \]
Note that since the magnetic field is parallel to one side of the loop, we should consider the effective area for torque calculation.

5. **Calculating the Torque**:
\[ \tau = 10 \cdot 0.01 \cdot 0.20 \cdot 1 = 0.02 \, \text{N m} \]

### Final Answers:
- (a) The force acting on the loop is \( 0.2 \, \text{N} \).
- (b) The torque acting on the loop is \( 0.02 \, \text{N m} \).