(i) \( 92: 54 \) 46 47 48 49 50 51 52 53 54 55 56 57 58 59 A coil is placed perpendicular to a uniform magnetic field of flux density (3T), and the magnetic flux passing 48. through the coil is ( \( \varphi \_m \) ), when the coil rotates by an angle of \( 60^{\circ} \). The magnitude of the magnetic flux density equals \( \qquad \) 0 T \( (3 \sqrt{ } 3) / 2 T \) 1.5 T \( 3 T \) NEXT QUESTION

To solve the problem, we need to determine the magnetic flux (\( \varphi_m \)) passing through the coil when it is rotated by an angle of \( 60^{\circ} \) in a uniform magnetic field with a flux density of \( 3 \, T \). ### Step 1: Understand the formula for magnetic flux The magnetic flux (\( \varphi \)) through a coil is given by the formula: \[ \varphi = B \cdot A \cdot \cos(\theta) \] where: - \( B \) is the magnetic flux density (in Tesla), - \( A \) is the area of the coil (in square meters), - \( \theta \) is the angle between the magnetic field and the normal to the surface of the coil. ### Step 2: Identify the known values From the problem: - \( B = 3 \, T \) - \( \theta = 60^{\circ} \) ### Step 3: Calculate the cosine of the angle We need to find \( \cos(60^{\circ}) \): \[ \cos(60^{\circ}) = \frac{1}{2} \] ### Step 4: Substitute the values into the formula Since the area \( A \) of the coil is not provided, we can express the magnetic flux in terms of \( A \): \[ \varphi = 3 \cdot A \cdot \frac{1}{2} = \frac{3A}{2} \] ### Step 5: Determine the magnitude of the magnetic flux density The problem asks for the magnitude of the magnetic flux density when the coil rotates. Since the magnetic flux density remains constant at \( 3 \, T \) regardless of the angle of rotation, the answer is simply the original magnetic flux density. ### Final Answer The magnitude of the magnetic flux density is: \[ 3 \, T \] Thus, the correct answer is \( 3 \, T \).