Question

A square loop of side L lies in the x-y plane in a magnetic field \( \vec{B} = B_0 (2\hat{i} + 3\hat{j} + 4\hat{k}) \), where \( B_0 \) is a constant. The magnetic flux through the loop is :

• A. \( \sqrt{29} B_0 L^2 \)
• B. \( 4 B_0 L^2 \)
• C. \( 3 B_0 L^2 \)
• D. \( 2 B_0 L^2 \)

Hint:

To find flux quickly, just multiply the area of the loop by the component of the magnetic field that is perpendicular to the loop's plane.
For an x-y loop, take the \( \hat{k} \) component of \( \vec{B} \). For a y-z loop, take the \( \hat{i} \) component of \( \vec{B} \).

Answer 1

The Correct Option is B

Concept:
• Magnetic flux (\( \Phi \)) represents the total number of magnetic field lines passing perpendicular to a given surface area.

• Mathematically, it is defined as the dot product of the uniform magnetic field vector (\( \vec{B} \)) and the area vector (\( \vec{A} \)) of the surface.

• \( \Phi = \vec{B} \cdot \vec{A} \).

• The area vector \( \vec{A} \) has a magnitude equal to the surface area and a direction strictly perpendicular (normal) to the surface plane.

Step 1: Determine the Area Vector of the square loop
The loop is a square with side length \( L \).
The magnitude of its area is \( A = L \times L = L^2 \).
The problem states that the loop lies in the x-y plane.
The vector strictly perpendicular to the x-y plane is the z-axis, represented by the unit vector \( \hat{k} \).
Therefore, the area vector is: \( \vec{A} = L^2 \hat{k} \).

Step 2: Calculate the Magnetic Flux using the Dot Product
The given magnetic field vector is \( \vec{B} = B_0 (2\hat{i} + 3\hat{j} + 4\hat{k}) \).
Apply the dot product formula for flux:
\[ \Phi = \vec{B} \cdot \vec{A} \]
\[ \Phi = \left[ B_0 (2\hat{i} + 3\hat{j} + 4\hat{k}) \right] \cdot \left[ L^2 \hat{k} \right] \]
In a dot product, orthogonal unit vectors yield zero (\( \hat{i} \cdot \hat{k} = 0 \) and \( \hat{j} \cdot \hat{k} = 0 \)), and parallel unit vectors yield one (\( \hat{k} \cdot \hat{k} = 1 \)).
\[ \Phi = B_0 L^2 (2(\hat{i} \cdot \hat{k}) + 3(\hat{j} \cdot \hat{k}) + 4(\hat{k} \cdot \hat{k})) \]
\[ \Phi = B_0 L^2 (0 + 0 + 4(1)) \]
\[ \Phi = 4 B_0 L^2 \]

Step 3: Conclusion
Only the z-component of the magnetic field contributes to the flux through a loop in the x-y plane.
The resulting magnetic flux is \( 4 B_0 L^2 \), making option (B) the correct answer.