Question

Figure shows a cross section of a large metal sheet carrying an electric current along its surface as shown in the figure. The current in a strip of width $dl$ is $kdl$, where $k$ is a constant. The magnetic field at a point $\text{P}$ distance $y$ from the metal sheet will be

• A. ${\mu }_{0}k$
• B. $\frac{{\mu }_{0}k}{2}$
• C. $\frac{{\mu }_{0}k}{4}$
• D. $\frac{3}{2}{\mu }_{0}k$

Answer 1

The correct option is B $\frac{{\mu }_{0}k}{2}$

Consider two strips $\text{A}$ and $\text{C}$ of the sheet situated symmetrically on the two sides of P.


Using right hand thumb rule, the magnetic field at $\text{P}$ due to strips $\text{A}$ and $\text{C}$ will be equal in magnitude due to symmetry (say $B$). The resultant magnetic field $B^{\prime }$ will be parallel to the sheet pointing leftwards as shown in the figure.

The net field on the opposite side of sheet at the same distance will also be $B^{\prime }$ but in opposite direction.

Let us construct a rectangular ampieran loop as shown below,

The net magnetic field on segment $\text{ad}$ and $\text{bc}$ will be zero.

Using, Ampere's circutal law for loop:

$\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_0{I}_{en}$

$\Rightarrow \left(B^{\prime }l\right)+\left(B^{\prime }l\right)={\mu }_0\left(k\right)l$

or, $2B^{\prime }l={\mu }_{0}kl$

Why this Question? Note: The magnetic field due to current carrying infinite sheet is independent of the distance of the point from the sheet.