Question

A square coil ABCD of side ' L ' is carrying a current in clockwise direction. A straight conductor carrying current \(\mathrm{I}_2\) (upward direction) is kept parallel to side AB at a distance \(\frac{\mathrm{L}}{3}\) in the plane of \(A B C D\). The net force on the coil ABCD is ( \(\mu_0=\) magnetic permeability)

• A. \(\frac{\mu _{0}\mathrm{I}_{1}\mathrm{I}_{2}}{3\pi}\)
• B. \(\frac{3\mu _{0}\mathrm{I}_{1}\mathrm{I}_{2}}{4\pi}\)
• C. \(\frac{3\mu _{0}\mathrm{I}_{1}\mathrm{I}_{2}}{8\pi}\)
• D. \(\frac{9\mu _{0}\mathrm{I}_{1}\mathrm{I}_{2}}{8\pi}\)

Hint:

Only the sides parallel to the wire feel a net force; the two forces are opposite and unequal, so subtract them.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The straight wire is parallel to side AB at distance \(L/3\). The forces on the two sides perpendicular to the wire cancel.

Step 2: Force on the near side
Side AB is at distance \(\dfrac L3\), so \(F_1=\dfrac{\mu_0I_1I_2L}{2\pi(L/3)}=\dfrac{3\mu_0I_1I_2}{2\pi}\).

Step 3: Force on the far side
Side CD is at distance \(L+\dfrac L3=\dfrac{4L}{3}\), so \(F_2=\dfrac{\mu_0I_1I_2L}{2\pi(4L/3)}=\dfrac{3\mu_0I_1I_2}{8\pi}\) in the opposite direction.

Step 4: Net force
\(F=F_1-F_2=\dfrac{3\mu_0I_1I_2}{2\pi}-\dfrac{3\mu_0I_1I_2}{8\pi}=\dfrac{9\mu_0I_1I_2}{8\pi}\).

Final Answer:
The net force on the coil is \(\dfrac{9\mu_0I_1I_2}{8\pi}\). \[ \boxed{\text{(D) }\dfrac{9\mu_{0}\mathrm{I}_{1}\mathrm{I}_{2}}{8\pi}} \]