Question

In the given figure, a large metal sheet of width $w$ carries a current $I$ (uniformly distributed in its width $w$. Find the magnetic field at point $P$, which lies in the plane of the sheet.

• A. $\frac{{\mu }_{0}I}{2\pi w}\left[ln\left(\frac{w}{a}\right)\right]$
• B. $\frac{{\mu }_{0}I}{2\pi w}\left[ln(1+\frac{w}{a})\right]$
• C. $\frac{{\mu }_0}{2\pi }\left[ln(1+\frac{w}{a})\right]$
• D. $\frac{{\mu }_{0}I}{4\pi w}\left[ln(1+\frac{w}{a})\right]$

Answer 1

The correct option is B $\frac{{\mu }_{0}I}{2\pi w}\left[ln(1+\frac{w}{a})\right]$

Given,
width of the strip $=w$, current $=I$



To find the field at point $\text{P}$ let us consider an infinitesimaly thin element of the sheet at a distance $x$ from point $\text{P}$ having thickness $dx$ as shown in the figure.

Field due to this small element at $\text{P}$ is

$dB=\frac{{\mu }_0\frac{I}{W}}{2\pi x}dx........\left(1\right)$

The magnetic field due to all the strips will be directed in a direction perpendicular to the sheet and pointing inwards.

So, integrating the eqn $\left(1\right)$ with proper limits we get total field at point $\text{P}$

$B=\int dB={\int }_a^{a+w}\frac{{\mu }_{0}I}{2\pi xw}dx$

$B=\frac{{\mu }_{0}I}{2\pi w}{\int }_a^{a+w}\frac{dx}{x}$

$=\frac{{\mu }_{0}I}{2\pi w}{\left[lnx\right]}_a^{a+w}$

$=\frac{{\mu }_{0}I}{2\pi w}\left[ln\frac{(a+w)}{a}\right]$

$\therefore B=\frac{{\mu }_{0}I}{2\pi w}\left[ln(1+\frac{w}{a})\right]$

Hence, option $\left(b\right)$ is the correct answer.