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Magnetic Fields due to Electric Current — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsMagnetic Fields due to Electric Current3 Marks
Question
Current of equal magnitude flows through two long parallel wires having separation of $1.35\ cm$. If the force per unit length on each of the wires in $4.76 \times 10^{-2} N$, what must be I ?
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Answer

Data : $I_1=I_2=I, s=1.35 \times 10^{-2} m , \frac{F}{l}=4.76 \times 10^{-2} N$
$F=\left(\frac{\mu_0}{4 \pi}\right) \frac{2 I_1 I_2 l}{s}=\left(\frac{\mu_0}{4 \pi}\right) \frac{2 I^2 l}{s}$
$\therefore I^2=\frac{F}{l} \frac{s}{2\left(\mu_0 / 4 \pi\right)}$
$\begin{aligned} & =\left(4.76 \times 10^{-2}\right) \frac{1.35 \times 10^{-2}}{2 \times 10^{-7}}=3.213 \times 10^3 \\ \therefore I & =\sqrt{32.13 \times 10^2}=56.68 A \end{aligned}$

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