A 2.0-m long conducting wire is formed into a square and placed in the horizontal x-y plane. A uniform magnetic field is oriented 30.0° above the horizontal with a strength of 9.0 T. What is the magnetic flux through the square?
The Correct Answer and Explanation is :
To calculate the magnetic flux (( \Phi_B )) through the square loop formed by the conducting wire, we can use the formula:
[
\Phi_B = B \cdot A \cdot \cos(\theta)
]
where:
- ( B ) is the magnetic field strength (in teslas),
- ( A ) is the area of the square (in square meters),
- ( \theta ) is the angle between the magnetic field and the normal to the surface of the square.
Step 1: Determine the Area of the Square
Given that the wire is 2.0 m long and is formed into a square, each side length (( L )) of the square can be calculated as follows:
[
L = \frac{2.0 \, \text{m}}{4} = 0.5 \, \text{m}
]
The area (( A )) of the square is:
[
A = L^2 = (0.5 \, \text{m})^2 = 0.25 \, \text{m}^2
]
Step 2: Identify the Magnetic Field Strength and Angle
The strength of the magnetic field (( B )) is given as 9.0 T. The magnetic field is oriented at 30.0° above the horizontal. The angle ( \theta ) we need is the angle between the magnetic field and the normal to the surface of the square. Since the square is in the horizontal x-y plane, the normal to this surface points vertically (90°). Therefore, the angle ( \theta ) is:
[
\theta = 90° – 30° = 60°
]
Step 3: Calculate the Magnetic Flux
Now we can plug these values into the magnetic flux formula:
[
\Phi_B = B \cdot A \cdot \cos(\theta)
]
Substituting the values:
[
\Phi_B = 9.0 \, \text{T} \cdot 0.25 \, \text{m}^2 \cdot \cos(60°)
]
Since ( \cos(60°) = 0.5 ):
[
\Phi_B = 9.0 \, \text{T} \cdot 0.25 \, \text{m}^2 \cdot 0.5
]
[
\Phi_B = 9.0 \cdot 0.125 = 1.125 \, \text{Wb} \quad (\text{Weber})
]
Conclusion
The magnetic flux through the square is ( \Phi_B = 1.125 \, \text{Wb} ). This calculation illustrates how the orientation of the magnetic field relative to the area affects the magnetic flux. A smaller angle (closer to 0°) increases the flux, while larger angles reduce it. The maximum flux occurs when the field is perpendicular to the area, which is when ( \theta = 0°).