Physics - Magnetism Question with Solution | TestHub

Let the two wires be at points $A$ and $B$. The distance between them is $AB=5$ cm.

Let the point be $P$. Given $PA=PB=r$ and $\angle APB=90^{\circ }$.

Using Pythagoras theorem in $△APB$:

$P{A}^2+P{B}^2=A{B}^2$

$r^2+r^2=5^2\Rightarrow 2r^2=25\Rightarrow r=\frac{5}{\sqrt{2}}$ cm $=\frac{5}{\sqrt{2}}\times 10^{-2}$ m.

The magnetic field at $P$ due to the first wire is $B_1=\frac{{\mu }_0{I}_1}{2\pi r}$.

The magnetic field at $P$ due to the second wire is $B_2=\frac{{\mu }_0{I}_2}{2\pi r}$.

Since the lines $PA$ and $PB$ are perpendicular, the magnetic fields ${\vec{B}}_1$ and ${\vec{B}}_2$ are also perpendicular to each other.

The net magnetic field at $P$ is $B=\sqrt{B_1^2+B_2^2}$.

$B=\frac{{\mu }_0}{2\pi r}\sqrt{I_1^2+I_2^2}$

Substituting the given values $I_1=4$ A and $I_2=3$ A:

$B=\frac{4\pi \times 10^{-7}}{2\pi \times \frac{5}{\sqrt{2}}\times 10^{-2}}\sqrt{4^2+3^2}$

$B=\frac{2\times 10^{-7}\times \sqrt{2}}{5\times 10^{-2}}\times 5$

$B=2\sqrt{2}\times 10^{-5}$ T.

Answer: $2\sqrt{2}\times 10^{-5}\mathrm{T}$