Solveeit Logo
Login

Question

Physics Question on Current electricity

A square loop ABCD, carrying a current I, is placed near and coplanar with a long straight conductor XY carrying a current I, as shown in figure. The net force on the loop will be

A

μ0I1I22π\frac{\mu_{0}I_{1}I_{2}}{2\pi}

B

μ0I1I2L2π\frac{\mu_{0}I_{1}I_{2}L}{2\pi}

C

2μ0I1I2L3π\frac{2\mu_{0}I_{1}I_{2}L}{3\pi}

D

2μ0I1I23π\frac{2\mu_{0}I_{1}I_{2}}{3\pi}

Answer

2μ0I1I23π\frac{2\mu_{0}I_{1}I_{2}}{3\pi}

Explanation

Solution

Force on arm AB due to current in conductor XY is F1=μ04π2I1I2L(L/2)=μ0I1I2πF_{1}=\frac{\mu_{0}}{4\pi} \frac{2I_{1}I_{2}L}{\left(L/2\right)}=\frac{\mu_{0}I_{1}I_{2}}{\pi} acting towards the wire in the plane of loop. Force on arm CD due to current in conductor XY is F2=μ04π2I1I2L(3L/2)=μ0I1I23πF_{2}=\frac{\mu_{0}}{4\pi} \frac{2I_{1}I_{2}L}{\left(3L/2\right)}=\frac{\mu_{0}I_{1}I_{2}}{3\pi} away from the wire the plane of loop. ∴\therefore\quad Net force on the loop =F1−F2=F_{1}-F_{2} =μ0I1I2π[1−13]=232μ0I1I2π\quad\quad=\frac{\mu_{0}I_{1}I_{2}}{\pi}\left[1-\frac{1}{3}\right]=\frac{2}{3} \frac{2\mu_{0}I_{1}I_{2}}{\pi}