Question

A square loop PQRS carrying a current of 6.0 A is placed near a long wire carrying 10 A as shown in figure. (a) Show that the magnetic force acting on the part PQ is equal and opposite to the part RS. (b) Find the magnetic force on the square loop. 

Answer 1

Given:
Current in the loop, i₁ = 6 A
Current in the wire, i₂ = 10 A 

Now, consider an element on PQ of width dx at a distance x from the wire. 

Force on the element is given by $$dF=\frac{{\mu }_0{i}_1{i}_2}{2\pi x}dx$$

Force acting on part PQ is given by 

$$F_{PQ}=\frac{{\mu }_0{i}_1{i}_2}{2\pi }\underset{1}{\overset{3}{\int }}\frac{dx}{x}$$ 

$$=2\times {10}^{-7}\times 6\times 10[\text{ ln }x{{]}_1}^3$$ 

$$=120\times {10}^{-7}\text{ ln }\left(3\right)N$$ 

$$\text{ Similarly },$$ 

$$F_{RS}=\frac{{\mu }_0{i}_1{i}_2}{2\pi }\underset{3}{\overset{1}{\int }}\frac{dx}{x}$$ 

$$=120\times {10}^{-7}\text{ ln }\left(\frac{1}{3}\right)$$ 

$$=-120\times {10}^{-7}\text{ ln }\left(3\right)N$$

Both forces are equal in magnitude, but they are opposite in direction.

(b) The magnetic field intensity due to wire on SP is given by

$$B=\frac{{\mu }_0{i}_2}{2\pi r}$$ 

Force on part SP is given by

 

$$F_{SP}=i_{1}Bl$$ 

$$=\frac{{\mu }_0{i}_1{i}_2}{2\pi }\left(\frac{l}{r}\right)$$ 

$$=\frac{{\mu }_0{i}_1{i}_2}{2\pi }\left(\frac{2}{1}\right)$$

(Towards right)

Force on part RQ is given by

$$F_{RQ}=i_{1}Bl$$
$$=\frac{{\mu }_0{i}_1{i}_2}{2\pi }\left(\frac{l}{r}\right)$$
$$=\frac{{\mu }_0{i}_1{i}_2}{2\pi }\left(\frac{2}{3}\right)$$

(Towards left)

Thus, the net force on the loop is given by

$$F_{net}=F_{SP}-F_{RQ}$$ 

$$=\frac{{\mu }_0{i}_1{i}_2}{2\pi }(\frac{2}{1}-\frac{2}{3})$$ 

$$=2\times {10}^{-7}\times 6\times 10\times \frac{4}{3}$$ 

$$=16\times {10}^{-6}$$  N

(Towards right)