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 find the torque on a circular loop placed in a magnetic field, we can use the formula for torque (\( \tau \)) on a current-carrying loop in a magnetic field:

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

Where:
- \( n \) = number of turns in the loop (for a single loop, \( n = 1 \))
- \( I \) = current in the loop (in Amperes)
- \( A \) = area of the loop (in square meters)
- \( B \) = magnetic field strength (in Teslas)
- \( \theta \) = angle between the normal to the loop and the magnetic field

In this case:
- The area \( A = 1 \, \text{cm}^2 = 1 \times 10^{-4} \, \text{m}^2 \) (since \( 1 \, \text{cm}^2 = 10^{-4} \, \text{m}^2 \))
- The current \( I = 10 \, \text{A} \)
- The magnetic field \( B = 0.1 \, \text{T} \)
- The angle \( \theta = 0^\circ \) (since the magnetic field is perpendicular to the plane of the loop, the normal to the loop is parallel to the magnetic field)

Now, substituting these values into the torque formula:

1. Calculate the torque:

\[ \tau = 1 \cdot 10 \cdot (1 \times 10^{-4}) \cdot 0.1 \cdot \sin(0) \]

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

\[ \tau = 1 \cdot 10 \cdot (1 \times 10^{-4}) \cdot 0.1 \cdot 0 \]

\[ \tau = 0 \]

Thus, the torque on the loop due to the magnetic field is \( 0 \, \text{N m} \).