Two long straight wires P and Q carrying equal current 10A each were kept parallel to each other at 5 cm distance. Magnitude of magnetic force experienced by 10 cm length of wire P is F$_1$. If distance between wires is halved and currents on them are doubled, force F$_2$ on 10 cm length of wire P will be:
Solution
$$
\begin{aligned}
& \text { Force per unit length between two parallel straight wires }=\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2}{2 \pi \mathrm{d}} \\\\
& \frac{\mathrm{F}_1}{\mathrm{~F}_2}=\frac{\frac{\mu_0(10)^2}{2 \pi(5 \mathrm{~cm})}}{\frac{\mu_0(20)^2}{2 \pi\left(\frac{5 \mathrm{~cm}}{2}\right)}}=\frac{1}{8} \\\\
& \Rightarrow \mathrm{F}_2=8 \mathrm{~F}_1
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Magnetic Effects of Current · Topic: Biot-Savart Law
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