Question

A square of side $L$ meters lies in the $x - y$ plane in a region where the magnetic field is given by$\overline B = {B_0}(2\widehat i + 3\widehat i + 4\widehat k)T$, where ${B_0}$ is constant. The magnitude of flux passing through the square is:
A. $8{B_0}{L^2}Wb$
B. $12{B_0}{L^2}Wb$
C. $4{B_0}{L^2}Wb$
D. $\sqrt {4 \times 29} {B_0}Wb$

Answer 1

Hint:Here we are going to apply the concept of magnetic flux which is defined as the number of magnetic lines of forces crossing the per unit area of a closed surface.

Complete step by step answer:
The magnetic flux is defined as the total number of magnetic fields of lines entering into any surface which is placed perpendicular to the magnetic field. It is the product of the perpendicular area which it enters and the average magnetic field strength.
We know that the square has a side of $L$ meters which lies in the $x - y$ plane in a region. And the magnetic field is given by $\overline B = {B_0}(2\widehat i + 3\widehat i + 4\widehat k)T$, where ${B_0}$ is constant.
Area of vector$ = \overrightarrow A = {L^2}\widehat k$, which is perpendicular to $x - y$ plane.
Magnetic field vector, $\overrightarrow B = {B_0}(2\widehat i + 3\widehat i + 4\widehat k)T$
$\phi = \overrightarrow {B.} \overrightarrow A $
$\phi = {B_0}(2\widehat i + 3\widehat i + 4\widehat k)T.{L^2}\widehat k$
$\phi = {B_0}(2{L^2}\widehat i.\widehat k) + 3{L^2}\widehat j.\widehat k + 4{L^2}\widehat k.\widehat k$
$\because \widehat i.\widehat k = 0,$ $\widehat j.\widehat k = 0$ and $\widehat k.\widehat k = 1$
$\therefore \phi = {B_0}4{L^2}$
$\therefore \phi = 4{B_0}{L^2}T$
Therefore, from the above explanation, the correct option is (C) $4{B_0}{L^2}Wb$.

Note:We should take care of the solid section, that is whether it is a square or rectangle, or else there be a chance of getting a wrong answer. We would have used the same method for electric flux as well.