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 Concept:
Magnetic flux (\(\Phi\)) is a scalar quantity that represents the total number of magnetic field lines passing through a surface.
Mathematically, it is the dot product of the magnetic field vector \(\vec{B}\) and the area vector \(\vec{A}\).
The area vector has a magnitude equal to the physical area and a direction perpendicular to the plane of the surface.
Crucially, only the component of the magnetic field that is normal (perpendicular) to the surface contributes to the magnetic flux.
Step 2: Key Formula or Approach:
Flux formula: \(\Phi = \vec{B} \cdot \vec{A}\).
Area of a square: \(Area = L^2\).
Area vector for a plane in the \(x-y\) plane: \(\vec{A} = (Area) \hat{k}\).
Step 3: Detailed Explanation:
The given square has a side length \(L\). Therefore, its area is \(L^2\).
Since the square lies in the \(x-y\) plane, the vector normal to this surface is the \(z\)-axis.
Thus, we can represent the area vector as \(\vec{A} = L^2 \hat{k}\).
The magnetic field vector is given as:
\[ \vec{B} = B_0 (2\hat{i} + 3\hat{j} + 4\hat{k}) = 2B_0\hat{i} + 3B_0\hat{j} + 4B_0\hat{k} \]
Now, perform the dot product calculation for \(\Phi\):
\[ \Phi = \vec{B} \cdot \vec{A} = (2B_0\hat{i} + 3B_0\hat{j} + 4B_0\hat{k}) \cdot (L^2 \hat{k}) \]
Expanding the dot product term by term:
\[ \Phi = (2B_0 L^2)(\hat{i} \cdot \hat{k}) + (3B_0 L^2)(\hat{j} \cdot \hat{k}) + (4B_0 L^2)(\hat{k} \cdot \hat{k}) \]
Using the orthogonality properties of unit vectors:
\(\hat{i} \cdot \hat{k} = 0\)
\(\hat{j} \cdot \hat{k} = 0\)
\(\hat{k} \cdot \hat{k} = 1\)
Substituting these into the equation:
\[ \Phi = 0 + 0 + 4B_0 L^2(1) \]
\[ \Phi = 4B_0 L^2 \]
The components of the magnetic field in the \(x\) and \(y\) directions are parallel to the plane of the square and thus do not penetrate it. Only the \(z\)-component contributes to the flux.
The final result is \(4 B_0 L^2\), matching option (D).
Step 4: Final Answer:
The magnetic flux is determined solely by the perpendicular component of the field, which is the \(\hat{k}\) component in this case, leading to the value \(4 B_0 L^2\).