Two parallel long wires carry currents $i_{1}$ and $i_{2}$ with $i_{1} i_{2}$. When the currents are in the same direction, the magnetic field midway between the wires is $10\mu T$. When the direction of $i_{2}$ is reversed, it becomes $40\ \mu T$. The ratio $i_{1}/i_{2}$ is

Question

Two parallel long wires carry currents $i_{1}$ and $i_{2}$ with $i_{1} > i_{2}$. When the currents are in the same direction, the magnetic field midway between the wires is $10\mu T$. When the direction of $i_{2}$ is reversed, it becomes $40\ \mu T$. The ratio $i_{1}/i_{2}$ is

Dr. Abubacker · Accepted Answer

(d) Initially when wires carry currents in the same direction as shown:
Magnetic field at mid point $O$ due to wires 1 and 2 are respectively
$B_{1} = \frac{\mu_{0}}{4\pi}.\frac{2i_{1}}{x}\bigotimes$
and $B_{2} = \frac{\mu_{0}}{4\pi}.\frac{2i_{2}}{x}\bigodot$
Hence net magnetic field at $O\ B_{net} = \frac{\mu_{0}}{4\pi} 	imes \frac{2}{x} 	imes (i_{1} - i_{2})$
$\Rightarrow 10 	imes 10^{- 6} = \frac{\mu_{0}}{4\pi}.\frac{2}{x}\left( i_{1} - i_{2} ight)\ \ \ldots(i)$
If the direction of $i_{2}$ is reversed then
$B_{1} = \frac{\mu_{0}}{4\pi}.\frac{2i_{1}}{x} \otimes$
and $B_{2} = \frac{\mu_{0}}{4\pi}.\frac{2i_{2}}{x} \otimes$
So $B_{net} = \frac{\mu_{0}}{4\pi}.\frac{2}{x}\left( i_{1} + i_{2} ight)$
$\Rightarrow 40 	imes 10^{- 6} = \frac{\mu_{0}}{4\pi}.\frac{2}{x}\left( i_{1} + i_{2} ight)\ \ldots(ii)$
Dividing equation (ii) by (i) $\frac{i_{1} + i_{2}}{i_{1} - i_{2}} = \frac{4}{1} \Rightarrow \frac{i_{1}}{i_{2}} = \frac{5}{3}$

Q. Two parallel long wires carry currents $i_{1}$ and $i_{2}$ with $i_{1} > i_{2}$. When the currents are in the same direction, the magnetic field midway between the wires is $10\mu T$. When the direction of $i_{2}$ is reversed, it becomes $40\ \mu T$. The ratio $i_{1}/i_{2}$ is

Explanation

Reviewed by Dr. Abubacker

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