\begin{tabular}{l} 10. A square loop of wire has a perimeter of 60 cm and is oriented such that two of its parallel \\ sides form a \( 30^{\circ} \) angle with the horizontal. A uniform horizontal magnetic field of magnitude \\ 0.2 T passes through the loop. (a) Calculate the magnetic flux through the loop. (b) Another \\ \hline\end{tabular}

To solve the problem, we need to calculate the magnetic flux through the square loop of wire. ### Given Data: - Perimeter of the square loop = 60 cm - Magnetic field strength (B) = 0.2 T - Angle (θ) between the magnetic field and the normal to the loop = \( 30^{\circ} \) ### Step 1: Calculate the side length of the square loop The perimeter \( P \) of a square is given by: \[ P = 4s \] where \( s \) is the side length. Given that the perimeter is 60 cm: \[ 60 = 4s \implies s = \frac{60}{4} = 15 \text{ cm} = 0.15 \text{ m} \] ### Step 2: Calculate the area of the square loop The area \( A \) of the square loop is given by: \[ A = s^2 = (0.15 \text{ m})^2 = 0.0225 \text{ m}^2 \] ### Step 3: Calculate the magnetic flux through the loop The magnetic flux \( \Phi \) through the loop is given by the formula: \[ \Phi = B \cdot A \cdot \cos(\theta) \] where: - \( B \) is the magnetic field strength, - \( A \) is the area of the loop, - \( \theta \) is the angle between the magnetic field and the normal to the surface of the loop. Since the loop is oriented such that two of its parallel sides form a \( 30^{\circ} \) angle with the horizontal, the angle \( \theta \) between the magnetic field (which is horizontal) and the normal to the loop (which is vertical) is: \[ \theta = 90^{\circ} - 30^{\circ} = 60^{\circ} \] Now, substituting the values into the flux formula: \[ \Phi = 0.2 \, \text{T} \cdot 0.0225 \, \text{m}^2 \cdot \cos(60^{\circ}) \] We know that \( \cos(60^{\circ}) = 0.5 \): \[ \Phi = 0.2 \cdot 0.0225 \cdot 0.5 \] \[ \Phi = 0.2 \cdot 0.01125 = 0.00225 \, \text{Wb} \] ### Final Answer: (a) The magnetic flux through the loop is \( \Phi = 0.00225 \, \text{Wb} \).