Figure shows a part of an electric circuit, PQRS is a rectangular loop made up of uniform wire in which PS = QR = 2 cm. The sides PQ and SR are sufficiently long as compared to the PS and QR. What will be the magnetic force per unit length acting on the wire SR due to the wire PQ if the ammeter reads 12 A?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 1.44\times10^{-3}\,\mathrm{N\,m^{-1}} \)
PQ and SR are long parallel wires, so the force per unit length between two parallel current-carrying wires applies. Here each of the two long sides carries the 12 A read by the ammeter, and they are \( d = PS = 2\,\mathrm{cm} = 0.02\,\mathrm{m} \) apart.
\( \displaystyle \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} \)
\( \displaystyle \frac{F}{L} = \frac{4\pi\times10^{-7}\times12\times12}{2\pi\times0.02} = \frac{2\times10^{-7}\times144}{0.02} \)
\( \displaystyle \frac{F}{L} = \frac{288\times10^{-7}}{2\times10^{-2}} = 1.44\times10^{-3}\,\mathrm{N\,m^{-1}} \)