The problem requires calculating the torque on a single-turn rectangular current loop situated in a magnetic field. The torque \((\tau)\) is determined using the formula: \(\tau = n \cdot I \cdot A \cdot B \cdot \sin(\theta)\)
where:
The loop dimensions are 20 cm by 10 cm. Converting to meters for area calculation: \(A = 0.2 \, \text{m} \times 0.1 \, \text{m} = 0.02 \, \text{m}^2\). The loop lies in the Y-Z plane, meaning its normal is along the X-axis. With the magnetic field along the positive Y-axis, \(\theta = 90^\circ\).
Substituting values into the torque equation:
\(\tau = 1 \times 5 \, \text{A} \times 0.02 \, \text{m}^2 \times 2 \times 10^{-3} \, \text{T} \times \sin(90^\circ) = 2 \times 10^{-4} \, \text{N} \cdot \text{m}\)
The direction of the torque is found using the right-hand rule. With the area vector along the X-axis and the magnetic field along the Y-axis, the torque is initially along the Z-axis. However, considering the anticlockwise current direction relative to the negative X-axis, the torque direction is reversed from the Y to Z curl, resulting in a direction along the negative Z-axis.
Therefore, the torque is \(2 \times 10^{-4} \, \text{N} \, \text{m}\) along the negative Z-direction.
A square Lamina OABC of length 10 cm is pivoted at \( O \). Forces act at Lamina as shown in figure. If Lamina remains stationary, then the magnitude of \( F \) is: 