Question

Four long, straight wires, each carrying a current of $5.0A$, are placed in a plane as shown in figure.
The points of intersection form a square of side $5.0cm$.
(a) Find the magnetic field at the centre $P$ of the square.
(b) $Q_1,Q_2,Q_3$ and $Q_4$ are points situated on the diagonals of the square and at a distance from $P$ that is equal to the length of the diagonal of the square. Find the magnetic fields at these points.

Answer 1

a) For each of the wire

Magnitude of magnetic field

$=\frac{{\mu }_{0}i}{4\pi r}(\sin {45}^{\circ }+\sin {45}^{\circ })=\frac{{\mu }_0\times 5}{4\pi \times \left(5/2\right)}\frac{2}{\sqrt{2}}$

For $AB\odot$ for $BC\odot$ for $CD\otimes$ and for $DA\otimes$.

The two $\odot$ and $2\otimes$ fields cancel each other. Thus $B_{net}=0$

b) At point $Q_1$

due to (1) $B=\frac{{\mu }_{0}i}{2\pi \times 2.5\times {10}^{-2}}=\frac{4\pi \times 5\times 2\times {10}^{-7}}{2\pi \times 5\times {10}^{-2}}=4\times {10}^{-5}\odot$

due to (2) $B=\frac{{\mu }_{0}i}{2\pi \times \left(15/2\right)\times {10}^{-2}}=\frac{4\pi \times 5\times 2\times {10}^{-7}}{2\pi \times 15\times {10}^{-2}}=\left(4/3\right)\times {10}^{-5}\odot$

due to (3) $B=\frac{{\mu }_{0}i}{2\pi \times (5+5/2)\times {10}^{-2}}=\frac{4\pi \times 5\times 2\times {10}^{-7}}{2\pi \times 15\times {10}^{-2}}=\left(4/3\right)\times {10}^{-5}\odot$

due to (4) $B=\frac{{\mu }_{0}i}{2\pi \times 2.5\times {10}^{-2}}=\frac{4\pi \times 5\times 2\times {10}^{-7}}{2\pi \times 5\times {10}^{-2}}=4\times {10}^{-5}\odot$

$B_{net}=[4+4+(4/3)+(4/3\left)\right]\times {10}^{-5}=\frac{32}{3}\times {10}^{-5}=10.6\times {10}^{-5}\approx 1.1\times {10}^{-4}T$