Question

A loop of area $4\,{{\text{m}}^{\text{2}}}$ is placed flat in the x-y plane. There is a constant magnetic field $4\widehat j$ in the region. Find the flux through the loop.
A. 2 unit
B. 4 unit
C. 6 unit
D. 0 unit

Answer 1

Use the formula for the magnetic flux through a loop in vector form. This formula gives the relation between the magnetic flux vector, magnetic field vector and area vector. Use the equation for the dot product of the two vectors.

Formula used:
The magnetic flux through a loop is given by
$\vec \phi = \vec B \cdot \vec A$ …… (1)
Here, $\vec \phi$ is the magnetic flux vector though the loop, $\vec B$ is the magnetic field vector and $\vec A$ is the area vector of the loop.
The formula for the for product of two vectors is
$\vec A \cdot \vec B = AB\cos \theta$ …… (2)
Here, $\vec A$ and $\vec B$ are the two vectors, $A$ and $B$ are the magnitudes of the vectors $\vec A$ and $\vec B$ respectively and $\theta$ is the angle between the vectors $\vec A$ and $\vec B$.

Complete step by step answer:
The loop of area $4\,{{\text{m}}^{\text{2}}}$ is placed flat in the x-y plane.
$A = 4\,{{\text{m}}^{\text{2}}}$
The area vector of an object in two dimensions is always perpendicular to the two dimensions and is
directed perpendicular in the third direction.
Since the loop is placed flat in the x-y plane, the area vector of the loop must be in a direction
perpendicular direction to x-y plane which is the Z-direction.
Hence, the area vector $\vec A$ of the loop is in the Z-direction.
$\vec A = \left( {4\,{{\text{m}}^{\text{2}}}} \right)$ $\widehat k$
The magnetic field vector $\vec B$ is in Y-direction.
$\vec B$ = $4\widehat j$
Determine the magnetic flux through the loop.
Substitute $4\widehat j$ for $\vec B$ and $\left( {4\,{{\text{m}}^{\text{2}}}} \right)$ $\widehat k$ for $\vec A$ in equation (1).
$\vec \phi$ = $4\widehat j$ (4 $m^2$) $\widehat k$
$\Rightarrow \vec \phi$ = $4\widehat j$ (4 $m^2$) $\cos \theta$
Here, $\theta$ is the angle between the area vector $\vec A$ and magnetic field vector $\vec B$.
The area vector is along the Z-axis and the magnetic field vector is along the Y-axis. The angle between
These two axes are $90^\circ$.
Substitute $90^\circ$ for $\theta$ in the above equation.
$\vec \phi$ = $4\widehat j$ (4 $m^2$) $\widehat k$ $\cos 90^\circ$
$\Rightarrow \vec \phi$ = $4\widehat j$ (4 $m^2$) $\widehat k$ $0$
$\Rightarrow \vec \phi$ = $0\,{\text{units}}$
Therefore, the magnetic flux through the loop is $0\,{\text{units}}$.

So, the correct answer is “Option D”.

Note:
One can also solve the dot product of the magnetic field vector and area vector using the information twidehat the dot product of the unit vector along Y-axis and the unit vector along Z-axis is zero.