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Magnetic Field due to a Current — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMagnetic Field due to a Current5 Marks
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 $5.0cm$.
  1. Find the magnetic field at the centre P of the square.
  2. $Q_1, Q_2, Q_3,$ and $Q_4$ are points situated on the diagonals of the liquare 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. For each of the wire
Magnitude of magnetic field
$=\frac{\mu_0\text{i}}{4\pi\text{r}}(\sin45^\circ+\sin45^\circ)=\frac{\mu_0\times5}{4\pi\Big(\frac{5}{2}\Big)}=\frac{2}{\sqrt2}$
For AB $\odot$ for BC $\odot$ for CD $\otimes$ and for DA $\otimes.$
The two $\odot$ and $2\otimes$ fields cancel each other. Thus $B_{net} = 0$
  1.  
At point $Q_1$,
Due to (1) $\text{B}=\frac{\mu_0\text{i}}{2\pi\times2.5\times10^{-2}}=\frac{4\pi\times5\times2\times10^{-7}}{2\pi\times5\times10^{-2}}=4\times10^{-5}\odot$
Due to (2) $\text{B}=\frac{\mu_0\text{i}}{2\pi\times\Big(\frac{15}{2}\Big)\times10^{-2}}=\frac{4\pi\times5\times2\times10^{-7}}{2\pi\times5\times10^{-2}}=\Big(\frac{4}{3}\Big)\times10^{-5}\odot$
Due to (3) $\text{B}=\frac{\mu_0\text{i}}{2\pi\times\Big(5+\frac{5}{2}\Big)\times10^{-2}}=\frac{4\pi\times5\times2\times10^{-7}}{2\pi\times15\times10^{-2}}=\Big(\frac{4}{3}\Big)\times10^{-5}\odot$
Due to (4) $\text{B}=\frac{\mu_0\text{i}}{2\pi\times2.5\times10^{-2}}=\frac{4\pi\times5\times2\times10^{-7}}{2\pi\times5\times10^{-2}}=4\times10^{-5}\odot$
$\text{B}_\text{net}=\Big[4+4+\Big(\frac{4}{3}\Big)+\Big(\frac{4}{3}\Big)\Big]\times10^{-5}$
$=\frac{32}{3}\times10^{-5}=10.6\times10^{-5}\approx1.1\times10^{-4}\text{T}$
At point $Q_2$,
Due to (1) $\frac{\mu_0\text{i}}{2\pi\times(2.5)\times10^{-2}}\odot$
Due to (2) $\frac{\mu_0\text{i}}{2\pi\times\Big(\frac{15}{2}\Big)\times10^{-2}}\odot$
Due to (3) $\text{B}=\frac{\mu_0\text{i}}{2\pi\times(2.5)\times10^{-2}}\otimes$
Due to (4) $\frac{\mu_0\text{i}}{2\pi\times\Big(\frac{15}{2}\Big)\times10^{-2}}\otimes$
At point $Q_3$,
Due to (1) $\frac{4\pi\times10^{-7}\times5}{2\pi\times\Big(\frac{15}{2}\Big)\times10^{-2}}=\frac{4}{3}\times10^{-5}\otimes$
Due to (2) $\frac{4\pi\times10^{-7}\times5}{2\pi\times\Big(\frac{5}{2}\Big)\times10^{-2}}=4\times10^{-5}\otimes$
Due to (3) $\frac{4\pi\times10^{-7}\times5}{2\pi\times\Big(\frac{5}{2}\Big)\times10^{-2}}=4\times10^{-5}\otimes$
Due to (4) $\frac{4\pi\times10^{-7}\times5}{2\pi\times\Big(\frac{15}{2}\Big)\times10^{-2}}=\frac{4}{3}\times10^{-5}\otimes$
$\text{B}_\text{net}=\Big[4+4+\Big(\frac{4}{3}\Big)+\Big(\frac{4}{3}\Big)\Big]\times10^{-5}$
$=\frac{32}{3}\times10^{-5}=10.6\times10^{-5}\approx1.1\times10^{-4}\text{T}$
For $Q_4​​​​​​​$,
Due to (1) $\frac{4}{3}\times10^{-5}\otimes$
Due to (2) $4\times10^{-5}\otimes$
Due to (3) $\frac{4}{3}\times10^{-5}\otimes$
Due to (4) $4\times10^{-5}\otimes$
$\text{B}_\text{net}=0$

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