Question

Two large metal sheets carry surface currents as shown in figure. The current through a strip of width $dl$ is $Kdl$, where $K$ is a constant. The magnetic field at point $\text{Q}$ will be

• A. ${\mu }_{0}K$
• B. $\frac{{\mu }_{0}K}{2}$
• C. zero
• D. $2{\mu }_{0}K$

Answer 1

The correct option is A ${\mu }_{0}K$

Considering a rectangular ampere loop passing through $\text{Q}$ as shown below,



Using Right-hand thumb rule the field due to the upper and lower metal sheet is towards right.

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

$\Rightarrow Bl={\mu }_0\left(Kl\right)$

$B={\mu }_{0}K$

Why this Question? The magnetic field at $\text{Q}$ is due to the superposition of magnetic fields due to the individual metal sheets. $B_{net}=B_1+B_2=\frac{{\mu }_{0}K}{2}+\frac{{\mu }_{0}K}{2}={\mu }_{0}K$