A uniform magnetic field passes through two areas, `A_(1)` and `A_(2)`. The angles between the magnetic field and the normals of areas `A_(1)` and `A_(2)` are 30.0° and 60.0°, respectively. If the magnetic flux through the two areas is the same, what is the ratio `A_(1)//A_(2)`?

Options

A. 0.577
B. 1
C. 0.816
D. 1.23

Answer

To solve the problem of finding the ratio \( \frac{A_1}{A_2} \) given that the magnetic flux through the two areas \( A_1 \) and \( A_2 \) is the same, we can follow these steps:

### Step 1: Understand the Magnetic Flux Formula
The magnetic flux \( \Phi \) through an area \( A \) in a magnetic field \( B \) is given by the formula:
\[ \Phi = B \cdot A \cdot \cos(\theta) \]
where \( \theta \) is the angle between the magnetic field and the normal (perpendicular) to the area.

### Step 2: Write the Flux Equations for Both Areas
For area \( A_1 \) with angle \( \theta_1 = 30^\circ \):
\[ \Phi_1 = B \cdot A_1 \cdot \cos(30^\circ) \]
For area \( A_2 \) with angle \( \theta_2 = 60^\circ \):
\[ \Phi_2 = B \cdot A_2 \cdot \cos(60^\circ) \]

### Step 3: Set the Fluxes Equal
Since the problem states that the magnetic flux through both areas is the same, we can set \( \Phi_1 \) equal to \( \Phi_2 \):
\[ B \cdot A_1 \cdot \cos(30^\circ) = B \cdot A_2 \cdot \cos(60^\circ) \]

### Step 4: Cancel Out Common Terms
We can cancel \( B \) from both sides of the equation (assuming \( B \) is not zero):
\[ A_1 \cdot \cos(30^\circ) = A_2 \cdot \cos(60^\circ) \]

### Step 5: Solve for the Ratio \( \frac{A_1}{A_2} \)
Rearranging the equation gives:
\[ \frac{A_1}{A_2} = \frac{\cos(60^\circ)}{\cos(30^\circ)} \]

### Step 6: Substitute the Values of Cosines
Using known values:
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \quad \text{and} \quad \cos(60^\circ) = \frac{1}{2} \]
Substituting these values into the ratio:
\[ \frac{A_1}{A_2} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} \]

### Step 7: Final Result
Thus, the ratio \( \frac{A_1}{A_2} \) is:
\[ \frac{A_1}{A_2} = \frac{1}{\sqrt{3}} \approx 0.577 \]