Question

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

• A. \( 5 B_0 L^2 \)
• B. \( 2 B_0 L^2 \)
• C. \( 3 B_0 L^2 \)
• D. \( 4 B_0 L^2 \)

Hint:

When a surface lies in a coordinate plane, only the component of the magnetic field perpendicular to that plane contributes to the magnetic flux. For a surface in the \( x-y \) plane, only the z-component (\( B_z \) or the coefficient of \( \hat{k} \)) is responsible for the flux, allowing you to quickly compute \( \Phi = B_z \times A \).

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem requires us to find the magnetic flux \( \Phi \) passing through a square of side \( L \) lying in the \( x-y \) plane, under a given uniform magnetic field \( \vec{B} \).

Step 2: Key Formula or Approach:
The magnetic flux \( \Phi \) is defined as the dot product of the magnetic field vector \( \vec{B} \) and the area vector \( \vec{A} \):
\[ \Phi = \vec{B} \cdot \vec{A} \]
Since the square lies in the \( x-y \) plane, its area vector is perpendicular to the plane and points along the z-axis (direction of unit vector \( \hat{k} \)):
\[ \vec{A} = A \hat{k} = L^2 \hat{k} \]

Step 3: Detailed Explanation:
Given:
- Magnetic field, \( \vec{B} = B_0(2\hat{i} + 3\hat{j} + 4\hat{k}) \)
- Area vector, \( \vec{A} = L^2 \hat{k} \)
Now, calculate the dot product:
\[ \Phi = \vec{B} \cdot \vec{A} \]
\[ \Phi = B_0(2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot (L^2 \hat{k}) \]
Since the dot product of perpendicular unit vectors is zero (\( \hat{i}\cdot\hat{k} = 0 \) and \( \hat{j}\cdot\hat{k} = 0 \)), and \( \hat{k}\cdot\hat{k} = 1 \):
\[ \Phi = B_0 \cdot 4 \cdot L^2 = 4 B_0 L^2 \]

Step 4: Final Answer:
The magnetic flux passing through the square is \( 4 B_0 L^2 \).