Understanding Magnetic Flux Through a Square Loop

This question asks us to calculate the magnetic flux ($\Phi$) passing through a square loop placed in a uniform magnetic field. We are given the side length of the loop, the strength of the magnetic field, and the angle the plane of the loop makes with the field. Magnetic flux is a fundamental concept in electromagnetism, representing the total amount of magnetic field lines passing through a specific area.

Key Concepts for Magnetic Flux Calculation

• Magnetic Flux ($\Phi$): Defined as the product of the magnetic field strength ($B$), the area ($A$) through which the field passes, and the cosine of the angle ($\theta$) between the magnetic field lines and the normal (perpendicular line) to the area. The formula is $\Phi = B A \cos(\theta)$.
• Area of a Square ($A$): For a square with side length '$a$', the area is calculated as $A = a^2$.
• Angle ($\theta$): It's crucial to use the angle between the magnetic field vector and the *normal* to the plane of the loop, not the angle the plane makes with the field. If the plane makes an angle $\alpha$ with the field, the normal makes an angle $\theta = 90^\circ - \alpha$ with the field.

Step-by-Step Solution for Magnetic Flux

Let's break down the calculation step-by-step:

1. Identify Given Values: We list the information provided in the question:

   • Side length of the square loop, $a = 1$ m
   • Resistance of the loop, $R = 1 \Omega$ (Note: Resistance is not needed to calculate flux).
   • Uniform magnetic field strength, $B = 0.5$ T
   • Angle between the plane of the loop and the magnetic field, $\alpha = 30^\circ$

2. Calculate the Area ($A$) of the Square Loop: The area of the square loop is found using its side length.

   Using the formula $A = a^2$:

   $ A = (1 \text{ m})^2 = 1 \text{ m}^2 $

3. Determine the Angle ($\theta$) between the Magnetic Field and the Normal: The question gives the angle between the loop's *plane* and the field ($\alpha = 30^\circ$). The magnetic flux formula requires the angle between the magnetic field ($B$) and the *normal* to the loop's plane.

   The normal vector is perpendicular to the plane. Therefore, the angle $\theta$ between the field and the normal is:

   $ \theta = 90^\circ - \alpha $

   $ \theta = 90^\circ - 30^\circ = 60^\circ $

4. Calculate the Magnetic Flux ($\Phi$): Now we can substitute the values into the magnetic flux formula $\Phi = B A \cos(\theta)$.

   $ \Phi = (0.5 \text{ T}) \times (1 \text{ m}^2) \times \cos(60^\circ) $

   We know that $\cos(60^\circ) = \frac{1}{2} = 0.5$. Substituting this value:

   $ \Phi = 0.5 \times 1 \times 0.5 $

   $ \Phi = 0.25 \text{ Weber (Wb)} $

Final Answer Analysis

The calculated magnetic flux through the square loop is $0.25$ Weber. This value directly corresponds to one of the options provided.

Comparing our result with the given options:

• Option 1: $\frac{\sqrt{3}}{4}$ Wb
• Option 2: $0.5$ Wb
• Option 3: $0.25$ Wb
• Option 4: Zero Wb

Our calculated value of $0.25$ Wb matches Option 3.