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Question

(A) Let the magnetic field be minimum at point $P$,at a distance $x$ from the first wire.The magnetic field due to the first wire at $P$ is $B_1 = \fr

Vedclass · Accepted Answer

$\frac{I_2}{I_1}=9$,anti-parallel

Vedclass · Answer

(A) Let the magnetic field be minimum at point $P$,at a distance $x$ from the first wire.The magnetic field due to the first wire at $P$ is $B_1 = \frac{\mu_0 I_1}{2 \pi x}$.The magnetic field due to the second wire at $P$ is $B_2 = \frac{\mu_0 I_2}{2 \pi (d-x)}$,where $d = 4 \, cm$.For the net magnetic field to be a non-zero minimum between the wires,the fields must be in opposite directions,which implies the currents must be anti-parallel.The net magnetic field is $B = |B_1 - B_2| = \frac{\mu_0}{2 \pi} |\frac{I_1}{x} - \frac{I_2}{d-x}|$.For $B$ to be minimum,the derivative $\frac{dB}{dx} = 0$.$\frac{d}{dx} (\frac{I_1}{x} - \frac{I_2}{d-x}) = 0 \Rightarrow -\frac{I_1}{x^2} - \frac{I_2}{(d-x)^2} = 0$.This implies $\frac{I_1}{x^2} = -\frac{I_2}{(d-x)^2}$. Since $I_1, I_2 > 0$,this confirms the currents are anti-parallel.Taking magnitudes: $\frac{I_2}{I_1} = \frac{(d-x)^2}{x^2}$.Given $d = 4 \, cm$ and $x = 1 \, cm$,we have $\frac{I_2}{I_1} = \frac{(4-1)^2}{1^2} = \frac{3^2}{1^2} = 9$.Thus,the ratio $\frac{I_2}{I_1} = 9$ and the currents are anti-parallel.

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