A circular coil of 25 turns and radius of 12 cm is placed in a uniform magnetic field of 0.5 T normal to the plane of coil. If the current in the coil is 5 A, then total torque experienced by the coil is

Options

A. 1.5Nm
B. 2.5Nm
C. 3.5Nm
D. zero

Explanation

To find the total torque experienced by the circular coil in a magnetic field, we can use the formula for torque (\( \tau \)) on a current-carrying coil in a magnetic field:

\[ \tau = n \cdot I \cdot A \cdot B \cdot \sin(\theta) \]

Where:
- \( n \) = number of turns in the coil
- \( I \) = current in the coil (in Amperes)
- \( A \) = area of the coil (in square meters)
- \( B \) = magnetic field strength (in Teslas)
- \( \theta \) = angle between the normal to the coil and the magnetic field

### Step 1: Identify and convert the given values
- Number of turns, \( n = 25 \)
- Radius of the coil, \( r = 12 \, \text{cm} = 0.12 \, \text{m} \)
- Current, \( I = 5 \, \text{A} \)
- Magnetic field, \( B = 0.5 \, \text{T} \)

### Step 2: Calculate the area of the coil
The area \( A \) of a circular coil is given by the formula:

\[ A = \pi r^2 \]

Substituting the radius:

\[ A = \pi (0.12)^2 = \pi (0.0144) \approx 0.04524 \, \text{m}^2 \]

### Step 3: Determine the angle \( \theta \)
Since the magnetic field is normal to the plane of the coil, the angle \( \theta = 0^\circ \).

### Step 4: Calculate the torque
Substituting the values into the torque formula:

\[ \tau = n \cdot I \cdot A \cdot B \cdot \sin(\theta) \]

Since \( \sin(0^\circ) = 0 \):

\[ \tau = 25 \cdot 5 \cdot 0.04524 \cdot 0 \]

Thus,

\[ \tau = 0 \]

### Conclusion
The total torque experienced by the coil is \( 0 \, \text{N m} \).

### Final Answer
The correct option is \( 0 \, \text{N m} \).

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